On Analytic Number Theory

Home , Analytic number theory

Lectures on Analytic Number Theory

By H. Rademacher

Tata Institute of Fundamental Research, Bombay 1954-55 Lectures on Analytic Number Theory

By H. Rademacher

Notes by K. Balagangadharan and V. Venugopal Rao

Tata Institute of Fundamental Research, Bombay 1954-1955 Contents

I Formal Power Series 1

1 Lecture 2

2 Lecture 11

3 Lecture 17

4 Lecture 23

5 Lecture 30

6 Lecture 39

7 Lecture 46

8 Lecture 55

II Analysis 59

9 Lecture 60

10 Lecture 67

11 Lecture 74

12 Lecture 82

13 Lecture 89

iii CONTENTS iv

14 Lecture 95

15 Lecture 100

III Analytic theory of partitions 108

16 Lecture 109

17 Lecture 118

18 Lecture 124

19 Lecture 129

20 Lecture 136

21 Lecture 143

22 Lecture 150

23 Lecture 155

24 Lecture 160

25 Lecture 165

26 Lecture 169

27 Lecture 174

28 Lecture 179

29 Lecture 183

30 Lecture 188

31 Lecture 194

32 Lecture 200 CONTENTS v

IV Representation by squares 207

33 Lecture 208

34 Lecture 214

35 Lecture 219

36 Lecture 225

37 Lecture 232

38 Lecture 237

39 Lecture 242

40 Lecture 246

41 Lecture 251

42 Lecture 256

43 Lecture 261

44 Lecture 264

45 Lecture 266

46 Lecture 272 Part I

Formal Power Series

1 Lecture 1

Introduction

In additive number theory we make reference to facts about addition in 1 contradistinction to multiplicative number theory, the foundations of which were laid by Euclid at about 300 B.C. Whereas one of the principal concerns of the latter theory is the deconposition of numbers into prime factors, additive number theory deals with the decomposition of numbers into summands. It asks such questions as: in how many ways can a given natural number be ecpressed as the sum of other natural numbers? Of course the decompostion into primary summands is trivial; it is therefore of interest to restrict in some way the nature of the summands(such as odd numbersor even numbers or per- fect squares) or the number of summands allowed. These are questions typical of those which will arise in this course. We shall have occasion to study the properties of -functions and their numerous applications to number theory, in particular theV theory of quadratic residues. Formal Power Series Additivenumber theory starts with Euler (1742). His tool was power series. His starting point was the simple relation xm. xn = xm+n by which multiplication of powers of x is pictured in the addition of exponents. He therefore found it expedient to use power series. Compare the situation in multiplicative number theory; to deal with the product n.m, one uses the equation nsms = (nm)s, thus paving the way for utilising Dirichlet series. While dealing with power series in modern mathematics one asks ques- 2 tions about the domain of convergence. Euler was intelligent enough not to ask this question. In the context of additive number theory power series are purely formal; thus the series 0! + 1! x + 2! x2 + is a perfectly good series in our ···

2 1. Lecture 3

theory. We have to introduce the algebra of formal power series in order to vindicate what Euler did with great tact and insight. 2 A formal power series is an expression a0 + a1 x + a2x + . Where the symbol x is an indeterminatesymbol i.e., it is never assigned a numer··· ical value. Consequently, all questions of convergence are irrelevant. Formal power series are manipulated in the same way as ordinary power series. We build an algebra with these by deﬁning addition and multiplication in the following way. If

∞ n ∞ n A = an x , B = bnx , = = Xn 0 Xn 0 ∞ n ∞ n we define A + B = C where C = cn x and AB = D where D = dn x , n=0 n=0 with the stipulation that we performP these operations in such a way thatP these equations are true modulo xN , whatever be N. (This reauirement stems from the fact that we can assign a valuation in the set of power series by defining ∞ n the order of A = anx to be k where ak is the first non-zero coefficient). n=0 Therefore cn and dnPmay be computed as for finite polynomials; then

cn = an + bn,

dn = a0bn + a1bn 1 + + an 1b1 + anb0. − ··· − A = B means that the two series are equal term by term, A = 0 means that all the coefficiants of A are zero. It is easy to verify that the following relations 3 hold: A + B = B + a AB = BA A + (B + C) = (A + B) + C A(BC) = (AB)C A(B + C) = AB + AC We summarise these facts by saying that the formal power series form a commutative ring. This will be the case when the coefficients are taken from such a ring, eg. the integres, real numbers, complex numbers. The ring of power series has the additional property that there are no divisors of zero (in case the ring of coefficients is itself an integrity domain), ie. if A, B = 0, either A = 0 or B = 0. We see this as follows: Suppose A = 0, B = 0. Let ak be the first non-zero coefficient in A, and b j the first non-zero coefficient ∞ n in B. Let AB = dnx ; then n=0 P dk+ j = a bk+ j + + ak 1b j+1 + akb j + ak+1b j 1 + + ak+ jb0 . ◦ ··· − − ··· 1. Lecture 4

In this expression the middle term is not zero while all the other terms are zero. Therefore dk+ j , 0 and so A.B , 0, which is a contradiction. From this property follows the cancellation law: If A , and A.B = A.C, then B = C. For, AB AC = A(B C). Since A , 0, B C◦ = or B = C. − − If the− ring of◦ coeﬃcients has a unit element so has the ring of power series. As an example of multiplication of formal power series, let, 4

A = 1 x and B = 1 + x + x2 + − ··· ∞ n A = anx , where a0 = 1, a1 = 1, and an = 0 for n 2, = − ≥ Xn 0 ∞ n B = bnx , where bn = 1, n = 0, 1, 2, 3,... = Xn 0 ∞ n C = cn x , where cn = a0bn + a1bn 1 + + anb0; = − ··· Xn 0 then

c0 = a0b0 = 1, cn = bn bn 1 = 1 1 = 0, n = 1, 2, 3,...; − − − so (1 x)(1 + x + x2 + ) = 1. − ··· We can very well give a meaning to inﬁnite sums and products in certain cases. Thus

A + A + = B, 1 2 ··· C C = D, 1 2 ··· both equations understood in the sense module xN , so that only a ﬁnite number N of A′ s or (C 1)′s can contribute as far as x . Let us apply− our methods to prove the identity:

1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4)(1 + x8) ··· ··· Let

C = (1 + x)(1 + x2)(1 + x4) ... (1 x)C = (1 x)(1 + x)(1 + x2)(1 + x4) ... − − = (1 x2)(1 + x2)(1 + x4) ... − = (1 x4)(1 + x4) ... − 1. Lecture 5

Continuing in this way, all powers of x on the right eventually disappear, and we have(1 x)C = 1. However we have shown that (1 x)(1+x+x2+ ) = 1, therefore (1 −x)C = (1 x)(1 + x + x2 + ), and by the− law of cancellation,··· C = 1 + x + x2−+ which− we were to prove.··· ··· This identity easily lends itself to an interpretation which gives an example 5 of the application of Euler’s idea. Once again we stress the simple fact that xn xm = xn+m. We have · 1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4)(1 + x8) ··· ··· This is an equality between two formal power series (one represented as a product). The coefficients must then be identical. The coefficient of xn on the right hand side is the number of ways in which n can be written as the sum of powers of 2. But the coefficient of xn on the left side is 1. We therefore conclude: every natural number can be expressed in one and only one way as the sum of powers of 2. We have proved that

1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4) ··· ··· If we replace x by x3 and repeat the whole story, modulo x3N , the coeﬃ- cients of these formal power series will still be equal:

1 + x3 + x6 + x9 + = (1 + x3)(1 + x2.3)(1 + x4.3) ··· ··· Similarly

1 + x5 + x2.5 + x3.5 + = (1 + x5)(1 + x2.5)(1 + x4.5) ··· ··· We continue indefinitely, replacing x by odd powers of x. It is permissible to multiply these infinitely many equations together, because any given power of x comes from only a finite number of factors. On the left appears

(1 + xk + x2k + x3k + ). ··· kYodd On the right side will occur factors of the form (1 + xN ). But N can be written uniquely as xλ.m where m is odd. That means for each N,1 + xN will occur once and only once on the right side. We would like to rearrange the factors to obtain (1 + x)(1 + x2)(1 + x3) ··· This may be done for the following reason. For any N, that part of the 6 formal power series up to xN is a polymial derived from a ﬁnite number of 1. Lecture 6

factors. Rearranging the factors will not change the polynomial. But since this is true for any N, the entire series will be unchanged by the rearrangement of factors. We have thus proved the identity

∞ (1 + xk + x2k + x3k + ) = (1 + xn) (1) ··· = kYodd Yn 1

∞ n This is an equality of two formal power series and could be written an x n=0 ∞ n = bn x . Let us ﬁnd what an and bn are. On the left we have P n=0 P (1 + x1.1 + x2.1 + x3.1 + )(1 + x1.3 + x2.3 + x3.3 + ) ··· ··· (1 + x1.5 + x2.5 + x3.5 + ) × ··· ··· xn will be obtainded as many times as n can be expressed as the sum of odd numbers, allowing repetitions. On the right side of (1), we have (1 + x)(1 + x2)(1 + x3) xn will be obtained as many times as n can be expressed as the sum of integers,··· no two of which are equal. an and bn are the number of ways in which n can be expressed respectively in the two manners just stated. But an = bn. Therefore we have proved the following theorem of Euler: Theorem 1. The number of representations of an integer n as the sum of different parts is the same as the number of representations of n as the sum of odd parts, repetitions permitted. We give now a diﬀerent proof of the identity (1).

∞ ∞ ∞ ∞ (1 + xn) (1 xn) = (1 xn)(1 + xn) = (1 x2n). = = − = − = − Yn 1 Yn 1 Yn 1 Yn 1 Again this interchange of the order of the factors is permissible. For, up to 7 any given power of x, the formal series is a polynomial which does not depend on the order of the factors.

∞ ∞ ∞ (1 + xn) (1 xn) = (1 x2n), = = − = − Yn 1 Yn 1 Yn 1 ∞ n ∞ 2n 1 ∞ 2n ∞ 2n (1 + x ) (1 x − ) (1 x ) = (1 x ). = = − = − = − Yn 1 Yn 1 Yn 1 Yn 1 1. Lecture 7

Now ∞ (1 x2n) , 0, and by the law of cancellation, we may cancel it from n=1 − both sidesQ of the equation obtaining,

∞ n ∞ 2n 1 (1 + x ) (1 x − ) = 1. = = − Yn 1 Yn 1 Multiplying both sides by

∞ 2n 1 2(2n 1) 3(2n 1) 1 + x − + x − + x − + ··· n=1 Y ∞ n ∞ 2n 1 ∞ 2n 1 2(2n 1) (1 + x ) 1 + x − 1 + x − + x − + ··· n=1 n=1 n=1 Y Y Y ∞ 2n 1 2(2n 1) = 1 + x − + x − + . ··· n=1 Y For the same reason as before, we may rearrange the order of the factors on the left.

∞ n ∞ 2n 1 2n 1 2(2n 1) (1 + x ) 1 + x − 1 + x − + x − + ··· n=1 n=1 Y Y ∞ 2n 1 2(2n 1) = 1 + x − + x − + . ··· n=1 Y However,

∞ 2n 1 2n 1 2(2n 1) 1 + x − 1 + x − + x − + = 1, ··· n=1 Y because we have shown that (1 x)(1 + x + x2 + ) = 1, and this remains true 2n 1 − ··· when x is replaced by x − . Therefore the above equation reduces to

∞ n ∞ 2n 1 2(2n 1) n 2n 3n (1 + x ) 5 1 + x − + x − + = 1 + x + x + x + ··· ··· n=1 n=1 n odd Y Y Y which is the identity (1). Theorem 1 is easily veriﬁed for 10 as follows:10, 1+9, 2+8, 3+7, 4+6, 8 1+2+7, 1+3+6, 1+4+5, 2+3+5, 1+2+3+4 are the unrestricted partitions. Par- titions into odd summands with repetitions are 1. Lecture 8

1+9, 3+7, 5+5, 1+1+1+7, 1+1+3+5, 1+3+3+3, 1+1+1+1+1+5, 1+1+1+1+3+3,1+1+1+1+1+1+1+3,1+1+1+1+1+1+1+1+1+1. We have ten partitions in each category. It will be useful to extend the theory of formal power series to allow us to find the reciprocal of the series a + a x + a x2 + where we assume that 0 1 2 ··· a0 , 0. (The coefficients are now assumed to form a field). If the series 1 b + b x + b x2 + = , 0 1 2 ··· a + a x + a x2 + 0 1 2 ··· we would have (a + a x + a x2 + )(b + b x + b x2 + ) = 1. This means 0 1 2 ··· 0 1 2 ··· that a0b0 = 1 and since a0 , 0, b0 = 1/a0. All other coefficients on the left vanish:

a0b1 + a1b0 = 0,

a0b2 + a1b1 + a2b0 = 0 ......

We may now find b1 from the first of these equations since all the a′ s and b0 are known. Then b2 can be found from the next equation, since b1 will then be known. Continuing in this, manner all the b′s can be computed by successively solving linear equations since the new unknown of any equation is always accompanied by aν , 0. The uniquely determined formal series b + b x + b x2 + is now called the reciprocal of a + a x + a x2 + (We 9 0 1 2 ··· 0 1 2 ··· can not invert if a0 = 0 since in that case we shall have to introduce negative exponentsand so shall be going out of our ring of power series). In view of this 1 definition it is meaningful to write = 1+ x+ x2 + since we have shown 1 x ··· − 1 that (1 x)(1 + x + x2 + ) = 1. Replacing x by xk, = 1 + xk + x2k + − ··· 1 xk ··· Using this expression, identity (1) may be written −

∞ 1 (1 + xn) 1 + xk + x2k + = . ··· 1 xk n=1 k odd k odd Y Y Y − For any N, 1 = 1 1 xk (1 xk) k odd Yk N − k odd,k N − ≤ ≤ Since this is true for any N, we may interchangeQ the order of factors in the entire product and get

∞ 1 = 1 (1 xk) (1 xk) n odd Y − k odd − Q 1. Lecture 9

Therefore, in its revised form identity (l) becomes:

∞ n = 1 (1 x ) n = − (1 x ) Yn 1 n odd − Q In order to determine in how many ways a number n can be split into k parts, Wuler introduced a parameter z into his formal power series. (The problem was proposed to Euler in St.Petersburgh: in how many ways can 50 be decomposed into the sum of 7 summands?). He considered such expression as (1 + xz)(1 + x2z) This is a formal power series in x. The coefficients of x are now polynomials··· in z, and since these polynomials form a ring they porvide an sdmissible set of coefficients. The product is not a formal power series in z 10 however. The coefficient of z for example, is an infinite sum which we do not allow.

(1 + xz)(1 + x2z)(1 + x3z) ··· = 1 + zx + zx2 + (z + z2)x3 + (z + z2)x4 + (z + 2z2)x5 + ··· = 1 + z(x + x2 + x3 + ) + z2(x3 + x4 + 2x5 + ) + ··· ··· ··· = 1 + zA (x) + z2A (x) + z3A (x) + (2) 1 2 3 ···

The expressions A1(x), A2(x), are themselves formal power series in x. They begin with higher and higher··· powers of x, for the lowest power of x 1+2+3+ +m m(m+1)/2 occurring in Am(x) is x ··· = x . This term arises by multiplying (xz)(x2z)(x3z) (xmz). The advantage in the use of the parameter z is that any power of x multiplying··· zm is obtained by multiplying m diﬀerent powers of x. Thus each term in Am(x) is the product of m powers of x. The z′ s therefore record the number of parts we have used in building up a number. N n Now we consider the ﬁnite product PN (z, x) (1 + zx ). ≡ n=1 = + (N) + 2 (N) + + PN(z, x) is a polynomial in z: PN (z, x) 1 zQA1 (x) z A2 (x) N (N) (N) = N(N+1)/2 ··· z AN (x), where AN (x) x . Replacing z by Zx, we have

N n+1 (1 + zx ) = PN (zx, x) = Yn 1 = 1 + zxA(N)(x) + z2x2A(N)(x) + 1 2 ··· So

N+1 (1 + zx)PN(zx, x) = 1 + zx PN (z, x), 1. Lecture 10

(1 + zx) 1 + zxA(N)(x) + + (zx)N A(N)(x) 1 ··· N = 1 + zxN+1 1 + A(N)(x) + z2A(N)(x) + 1 2 ··· We may now compare powers of z on both sides since these are polynomi- 11 als. Taking zk, k N, we have ≤ k (N) + k (N) = (N) + N+1 (N) x Ak (x) x Ak 1(x) Ak (x) x Ak 1(x); − − (N) k = (N) k N+1 k Ak (x)(1 x ) ak 1(x)x 1 x − , − − − k (N) x N+1 k (N) A (x) = 1 x − A (x), k 1 xk − k 1 − − xk A(N)(x) A(N) (x) (mod xN ). k ≡ 1 xk k 1 − − From this recurrence relation we immediately have x A(N)(x) (mod xN ), 1 ≡ 1 x − x x2 A(N)(x) · (mod xN ) 2 ≡ (1 x)(1 x2) − − x3 (mod xN ) ≡ (1 x)(1 x2) − − ...... xk(k+1)/2 A(N) (mod xN ) k ≡ (1 x)(1 x2) (1 xk) − − ··· − Hence

∞ zx z2 x3 z3x6 + z n + + + (1 x ) 1 2 2 3 = ≡ 1 x (1 x)(1 x ) (1 x)(1 x )(1 x ) Yn 1 − − − − − − + (mod xN ) ··· Lecture 2

In the last lecture we proved the identity: 12

∞ k ∞ k (1 + zx ) = z Ak(x), (1) = = Yn 1 Xk 0 where xk(k+1)/2 A (x) = (2) k (1 x)(1 x2) (1 xk) − − ··· − We shall look upon the right side of (1) as a power series in x and not as a power-series in z, as otherwise the inﬁnite product on the left side would have no sense in our formalism. Let us inerpret (1) arithmetically. If we want to decompose m into k summands, we have evidently to look for zk and then for xm, and the coeﬃcient of zk xm on the right side of (1) gives us exactly what we want. We have

1 ∞ ∞ ∞ = n1 2n2 knk 2 k x x x (1 x)(1 x ) (1 x ) = = ··· = − − ··· − nX1 0 nX2 0 nXk 0 = ∞ (k) m pm x , = Xm 0 (k) = say, with p0 1. Therefore m occurs only in the form

m = n + 2n + + kn , n 0, 1 2 ··· k j ≥ (k) and pm tells us how often m can be represented by k dﬀerent summands (with possible repetitions). On the other hand the coeﬃcient of xm on the left-side

11 2. Lecture 12

of (1) gives us the number of partitions of m into summands not exceeding k. Hence,

Theorem 2. m can be represented as the sum of k different parts as often as 13 k(k + 1) m can be expressed as the sum of parts not exceeding k (repetition − 2 being allowed). (In the first the number of parts is fixed, in the second, the size of parts). In a similar way, we can extablish the identity

1 ∞ = zk n Bk(x), (3) n∞=1(1 zx ) = − Xk 0 Q with B0 = 1, which again can be interpreted arithmetically as follows. The left side is

∞ ∞ ∞ (zx)n1 (zx2)n2 (zx3)n3 = = = ··· nX1 0 nX2 0 nX3 0 and xk B (x) = (4) k (1 x)(1 x2) (1 xk) − − ··· − The left-side of (3) gives m with the representation

m = n + 2n + 3n + 1 2 3 ··· i.e., as a sum of parts with repetitions allowed. Exactly as above we have: Theorem 3. m can be expressed as the sum of k parts (repetitions allowed) as often as m k as the sum of parts not exceeding k. − We shall now consider odd summands which will be of interest in connex- ion with -function later. As earlier we can establish the identity V

∞ k (1 + zxV) = z Ck(x) (5) odd k=0 VY X with the provide that C (x) = 1. The trick is the same. One studies temporatily 14 V ◦ 2 a truncated aﬀair (1 + zxV), replaces z by zx and evaluates Ck(x) as in =1 V Lecture 1. This wouldQ be perfectly legitimate. However one could proceed as 2. Lecture 13

Euler did - this is not quite our story. Multiplying both sidesby 1 + zx2, we have ∞ k 2 ∞ k 2k z Ck(x) = (1 + zx ) z x Ck(x). = = Xk 0 Xk 0 Now compare powers of z on both sides - and this was what required some 1+3+ +(2k 1) k2 extra thought. Ck(x) begins with x ··· − = x ; in fact they begin with later and later powers of x and so can be added up. We have

C0 = 1, 2k 2k 1 Ck(x) = x Ck(x) + x − Ck 1(x), k > 0, − 2k 1 x − or Ck(x) = Ck 1(x) 1 x2k − − from this recurrence relation we obtain x C (x) = , 1 1 x2 − x3 x4 C (x) = C (x) = , 2 1 x4 1 (1 x2)(1 x4) − − − x5 x9 C (x) = C (x) = , 3 1 x6 2 (1 x2)(1 x4)(1 x6) − − − − ...... xk2 C (x) = , k (1 x2)(1 x4) (1 x2k) − − ··· − carrying on the same rigmarole. Now note that all this can be retranslated into something. Let us give the number theoretic interpretation. The coefficient of zk xm 15 gives the number of times m can be expressed as the sum of k different odd ffi 1 summands. On the other hand, the coe cients in the expansion of (1 x2) (1 x2k) give the decomposition into even summands, with repetitions. Hence,− ··· − Theorem 4. misthesumofk different odd parts as often as m k2 is the sum m− k2 of even parts not exceeding 2k, or what is the same thing, as −2 is the sum of parts not exceeding k. (since m and k are obviously of the same parity, it m k2 follows that −2 is an integer). Finally we can prove that

1 ∞ k = z Dk(x) (6) odd(1 zxV) = V − Xk 0 Q 2. Lecture 14

Replacing z by zx2, we obtain

xk D (x) = , k (1 x2) (1 x2k) − ··· − leading to the Theorem 5. m is the sum of k odd parts as often as m k is the sum of even m k − m k parts not exceeding 2k, or 2− is the sum of even parts not exceeding k. ( 2− again is integral). Some other methods Temporarily we give up power series and make use of graphs to study partitions. A partition of may be represented as an array of dots, the number of dots in a row being equalN to the magnitude of a summand. Let us arrange the summands according to size. For instance, let us consider a partition of 18 into 4 diﬀerent parts 16

If we read the diagram by rows we get the partiton 18=7+5+4+2. On the other hand reading by columns we have the pertition 18= 4+4+3+3+2+1+1. In general it is clear that if we represent a partition of n into k parts graphically, then reading the graph vertically yields a partition of n with the largest part k, and conversely. This method demonstrates a one-to-one correspondence between partitions of n with k parts and partitions sees that the number of partitions of n with largest part k is equal to the number of partitions of n k into parts not exceeding k. − 2. Lecture 15

Draw a diagonal upward starting from the last but one dot in the column on theextremeleft. All thedotsto the rightof this diagonalconstitute a partition of 12 into 4 parts. For each partition of 18 into 4 different parts there corresponds 4.3 = thus a partition of 18 2 12 into parts. This process works in general for a 17 partition of n with k di−fferent parts. If we throw away the dots on and to the left of the diagonal (which is drawn from the last but one point from the bottom in order to exsure that the number of different parts constinues to be exactly k), + + + + = k(k 1) we are left with a partition of n (1 2 3 (k 1)) n 2− . This partition has exactly k parts because− each row is··· longer− by at least− one dot than the row below it, so an entire row is never discarded. Conversely, starting with k(k 1) a partition of n 2− into k parts, we can build up a unique partition of n into k different parts.− Add 1 to the next to the smallest part, 2 to the next longer, 3 to the next and so on. This one-to-one correspondence proves that the number of ff k(k 1) partitions of n into k di erent parts equals the number of partitions of n 2− into k parts. − We can prove graphically that the number of partitons of n into k odd summands is the same as the number of partitions of n k2 into even summands not exceeding k. The last row of the −

diagram contains at least one dot, the next higher at least three, the one above at least ﬁve, and so on. Above and on the diagonal there are 1 + 3 + 5 + + (2k 1) = k2 dots. When these are removed, an even number of dots is ···left in each− row, althogether adding up to n k2. This proves the result. − Theorem 1 can also be proved graphically, although the proof is not quite 18 as simple. The idea of the proof is exampliﬁed by considering the partitons of 35. We have

35 = 10 + 8 + 7 + 5 + 4 + 1 = 5 2 + 1 8 + 7 + 5 + 1 4 + 1 × × × = 5(2 + 1) + 7 1 + 1(8 + 4 + 1) × 2. Lecture 16

7 + 5 + 5 + 5 + 1 + + 1  ···   13 times    Thus to each unrestricted partition of| 35{z we} can make correspond a partition into add summands with possible repetitions. Conversely 7 1 + 5 3 + 1 13 = 7 1 + 5(1 + 2) + 1(23 + 22 + 20) × × × × = 7 + 5 + 10 + 8 + 4 + 1. Now consider the following diagram

2 4 6 3

13 times

Each part is represented by a row of dots with the longest row at ehe top, second longest next to the top, etc. The oddness of the parts allows uo to place the rows symmetrically about a central vertical axis. Now connect the 19 dots in the following way. Connect the dots on this vertical axis with those on the left half of the top row. Then connect the column to the right of this axis to the other half of the top row. We continue in this way as indicated by the diagram drawing right angles first on one side of the centre and then on the other. We now interpret this diagram as a new partition of 35 each part being represented by one of the lines indicated. In this way we obtain the partition 20+6+4+3+2of35intodifferentparts. It can be provedthat this method works in general. That is, to prove that given a partition of n into odd parts, this method transforms it into a unique partition of n into distinct parts; conversely, given a partation into distinct parts, the process can be reversed to find a unique partition into odd parts. This establishes a one-to-one correspondence between the two sorts of partitions. This proves our theorem. Lecture 3

∞ k The series z Ak(x) that we had last time is itself rather interesting; the Ak(x) 20 k=0 have a queerP shape: xk(k 1)/2 A (x) = − k (1 x)(1 x2) (1 xk) − − ··· − Such series are called Euler series. Such expressions in which the factors in the denominator are increasing in this way have been used for wide generalisations of hypergeometric series. Euler indeed solved the problem of computing the coefficients numerically. The coefficient of zk xm is obtained by expanding 1 (1 x) (1 xk ) as a power series. This is rather trivial if we are in the field of complex− ··· numbers,− since we can then have a decomposition into partial fractiions. Euler did find a nice sort of recursion formula. There is therefore a good deal to be said for a rather elementary treatment. We shall, however, proceed to more important discussions the problem of unrestricted partitions. Consider the infinite product (this is justifiable modulo xN )

∞ 1 ∞ ∞ = xmn 1 xm m=1 m=1 n=0 Y − Y X ∞ ∞ ∞ = xn1 x2n2 x3n3 = = · = ··· nX1 0 nX2 0 nXj 0 = 1 + x + 2x2 + ··· ∞ n = 1 + pn x (1) = Xn 1

17 3. Lecture 18

n What does pn signify? pn appeared in collecting the term x . Following 21 Euler’s idea of addition of exponents, we have n = n + 2n + 3n + 4n + n 0, (2) 1 2 3 4 ··· j ≥ so that pn is the number of solutions of a finite Diophantine equation (since the right side of (2) becomes void after a finite stage) or the number of ways in which n can be expressed in this way, or the number of unrestricted partitions. Euler wrote this as 1 = ∞ n m p(n)x , (3) ∞ (1 x ) m=1 n=0 − X with the provide that p(0)Q= 1. We want to find as much as possible about p(n). Let us calculate p(n). Expanding the product,

∞ (1 xn) = (1 x)(1 x2)(1 x3) − − − − ··· n=1 Y = 1 x x2 + x5 + x7 x12 x15 ++ − − − − −−··· (Note Euler’s skill and patience; he calculated up to xn and found to this surprise that the coefficients were always 0, 1, two positive terms followed by two negative terms). We want to find the law± of exponents, as every sensible man would. Writing down the first few coeffieicnts and taking differences, we have 0 1 2 5 7 12 15 22 26 1 1 3 2 5 3 7 4 the sequence of odd numbers interspersed with the sequence of natural numbers. Euler forecast by induction what the general power would be as follows. 22

7 2 0 1 5 12 22 5 2 1 4 7 10 − − 3 333 3 Write down the coefficients by picking up 0, 1 and every other alternate term, and continue the row towards the left by putting in the remaining coefficients. Now we find that the second differences have the constant value 3. But an arithmetical progression of the second order can be expressed as a polynomial of the second degree. The typical coefficient will therefore be given by an expression of the form 3. Lecture 19

aλ2 + bλ + c a(λ + 1)2 + b(λ + 1) + c a(λ + 2)2 + b(λ + 2) + c

a(2λ + 1) + b a(2λ + 3) + b

2 a (the constant second difference) = = = = = 1 Hence 2a 3 or a 3/2. Taking λ 0 we find that c 0 and b 2 , ffi λ(3λ 1) − so that the general coe cient has the form 2− . Observing that when λ is λ(3λ 1) λ(3λ+1) ffi λ(3λ 1)/2 λ changed to λ, 2− becomes 2 , the coe cient of x − is ( ) , and hence − − ∞ n ∞ λ λ(3λ 1)/2 (1 x ) = ( ) x − , (4) − − n=1 λ= Y Y−∞ which is Euler’s theorem. λ(3λ 1) This sequence of numbers 2− played a particular role in the middle ages. They are called pentagonal numbers and Euler’s theorem is called the pentagonal numbers theorem. We have the so-called triangular numbers:

1 3 6 10 15 2 3 4 5 1 1 1

where the second diﬀerences are all 1; the square-numbers 23

1 4 9 16 25 3 5 7 9 2 2 2

for which the second diﬀerence are always 2; and so on.

The triangular numbers can be represented by dots piled up in the form of equilateral triangles; the square numbers by successively expanding squares. 3. Lecture 20

The pentagons however do not ﬁt together like this. We start with one pen- tagon; notice that the vertices lie perspectively along rays through the origin. So take two sides as basic and magnify them and add successive shelves. The second diﬀerences now are always 3:

1 5 12 22

4 7 10 3 3

In general we can have r-gonal numbers where the last diﬀerence are all r 2. − We go back to equation (4): 24

∞ m ∞ λ λ(3λ 1)/2 (1 x ) = ( ) x − − − m=1 λ= Y X−∞ It is quite interesting to go into the history of this. It appeared in Euler’s Introductio in Analysin Infinitorum, Caput XVI, de Partitio numerorum, 1748 (the first book on the differential and integral calculus). It was actually discovered earlier and was mentioned in a paper communicated to the St. Petersburgh Academy in 1741, and in letters to Nicholas Bernoulli (1742) and Goldbach (1743). The proof that bothered him for nine years was first given in a letter dated 9th june 1750 to Goldvach, and was printed in 1750. The identity (4) is remarkable; it was the first time in history that an identity belonging to the -functions appeared (later invented and studied systemati- cally by Jocobi).V The interesting fact is that we have a power-series in which the exponentsare of the second degree in the subscripts. The -functions have a representation as a series and slso as an infinite porduct. V The proof of identity (4) is quite exciting and elementary. By using dis- tributivity we break up the product

(1 x)(1 x2)(1 x3)(1 x4) − − − − ··· in the following way:

(1 x)(1 x2)(1 x3)(1 x4) = 1 x (1 x)x2 (1 x)(1 x2)x3 − − − − ··· − − − − − − − 3. Lecture 21

(1 x)(1 x2)(1 x3)(1 x4) (1 x) (1 x4)x5 − − − − − − ··· − −··· which may be re-arranged, opening ﬁrst parenthesis, as

1 x x2 (1 x2)x3 (1 x2)(1 x3)x4 (1 x2)(1 x3)(1 x4)x5 − −E − − M − − − RR − − − J − EE MM RRR JJ EE MMM RRR JJ EE MM RRR JJ EE MM RRR JJ JJJ +x3 +(1 x2)x4 +(1 x2)(1 x3)x5 − − − So 25

1 x x2 + x5 + x7(1 x2) + x9(1 x2)(1 x3) + − − − − − ··· = 1 x x2 + x5+ 27 +(1 x3)x9 +(1 x3)(1 x4)x11 − − :: − MM − −CC : MMM CC :: MM CC :: MM CC M CC x9 (1 x3)x11 − − − 1 x x2 + x5 + x7 x12(1 x3)x15 (1 x3)(1 x4)x18 − − − − − − − −··· When this is continued, we get some free terms at the beginning followed by a typical remainder

(1 xk)xm + (1 xk)(1 xk+1)xm+k + (1 xk)(1 xk+1)(1 xk+2)xm+2k, − − − − − − which may be rearranged into

xm + (1 xk+1)xm+k + (1 xk+1)(1 xk+2)xm+2k xm+k (1 xk+1)xm+2k (*) − − − − − − = xm xm+2k+1 (1 xk+1)xm+3k+2 (1 xk+1)(1 xk+2) − − − − − − xm+4k+3 (**) −··· We have two free terms with opposite signs at the beginning. In (*) the difference between exponents in successive terms is k, while in (**) this increases to k + 1; this difference is in both cases the exponent of x in the first factor. The remainder after the free terms begine with , so that the sequence − of signs is + ++ This process perpetuates itself and the question 26 remains which−− powers−−··· actually appear. It is sufficient to mark down a scheme for the exponents which completely characterises the expansion. The scheme is illustrated by what follows. 3. Lecture 22

01234 5 6 7 8 910111213141516 23 4 5 6 7 8 9101112131415 5 7 9 11 13 15 17 19 21 23 25 27 29 31 3 4 5 6 7 8 91011121314 12 15 18 21 24 27 30 33 36 39 42 45 4 5 6 7 8 910111213 22 26 30 34 38 42 46 50 54 58 5 6 7 8 9101112 35 40 45 50 55 60 65 70 6 7 8 91011 51 57 63 69 75 81 We write down the sequence of natural numbers in a row; the sequence less the ﬁrst two membere is repeated in a parallel row below leaving out the ﬁrst three placess at the beginning. Adding up we get

5 7 9 11 ...... ,

below which is placed the original sequence less the ﬁrst three members, again translating the whole to the right by two places. We again add up and repeat the procedure. A typical stage in the procedure is exhibited below.

m m + k m + 2k m + 3k m + 4k m + 5k k + 1 k + 2 k + 3 k + 4 k + 5 m + 2k + 1 m + 3k + 2 m + 4k + 3 m + 5k + 4 m + 6k + 5

The free indices then appear successively as 27

2 + 3 = 5 3 + 4 + 5 = 12 3 + 4 = 7 4 + 5 + 6= 15, and in general: λ(3λ 1) λ + (λ + 1) + + (2λ 1) = − , ··· − 2 λ(3λ + 1) (λ + 1) + (λ + 2) + + 2λ = , ··· 2 which are the only exponents appearing. We thus have

∞ n ∞ λ λ(3λ 1)/2 (1 x ) = ( ) x − − − n=1 λ= Y X−∞ Lecture 4

In the last lecture we proved the surprising theorem on pentagonal numbers: 28

∞ m ∞ λ λ(3λ 1)/2 (1 x ) = ( ) x − (1) − − m=1 λ= Y X−∞ We do not need these identities for their own sake, but for their applications to number theory. We have the same sort of power-series on both sides; let us compare the coefficients of xn. Onthe leftside n appearsas the sum of different exponents. But in contradiction to previous situations, the coefficients appear with both positive and negative signs, so that when we collect the terms there may be cancellations. There are gaps in the powers that appear, but among those which appear with non-zero coefficients, we have a pair of positive terms followed by a pair of negative terms and vice versa. In most cases the coefficients are zero; this is because of cancellations, so that roughly terms with positive and negative signs are in equal number. A positive sign appears if we multiply an even number of times. otherwise a negative sign. So an even number of different summands is as frequent generally as an odd number. Hence the following theorem: The number of decompositions of n into an even number of different parts is the same as the number of decompositions into an odd number, with the exception that there is a surplus of one sort or the other if n is a pentagonal number of the form λ(3λ 1)/2. Before proceeding further− let us examine a number of concrete instances. Take 6 which is not a pentagonal number. The partitions are 6, 1 + 5, 2 + 4, 29 1 + 2 + 3, so that there are two decompositions into an even number of different parts, and two into an odd number. Next take 7, which is a pentagonal number, = λ(3λ+1) = 7 2 with λ 2. We can actually foresee that the excess will be in the even partitions. The partitions are 7, 1 + 6, 2 + 5, 3 + 4, 1 + 2 + 4. Take 8 which

23 4. Lecture 24

again is not pentagonal. We have three in each category: 8, 1 + 7, 2 + 6, 3 + 5, 1 + 2 + 5, 1 + 3 + 4. This is a very extraordinary property of pentagonal numbers. One would like to have a direct proof of this. A proof is due to Fabian Franklin (Comptes Rendus, Paris. 1880), a pupil of the famous Sylvester. The proof is combinatorial. We want to establish a one-one correspondence between partitions containing an even number of summands and those containing an odd number - except for pentagonal numbers. Consider a partition with the summands arranged in increasing order, each summand being denoted by a horizontal row of dots. Mark speciﬁcally the ﬁrst row,

with r dots, and the last slope, with s dots i.e., points on or below a segn- ment starting from the dot on the extreme right of the last row and inclined at 45◦ (as in the diagram). We make out two cases. 1. s < r. Transfer the last slope to a position immediately above the ﬁrst row. The diagram is now as shown below:

The uppermost row is still shorter than the others. (because in our case 30 s < r). By this procedure the number of rows is changed by 1. This establishes the one-one correspondence between partition of the ‘odd’ type and ‘even’ type. 4. Lecture 25

2. s r. As before consider the ﬁrst row and the last slope. ≥

Take the uppermost row away and put it parallel to the last slope. This diminishes the number of rows by 1, so that a partition is switched over from the ‘even’ class to the ‘odd’ class or conversely. Therefore there exists a one-one correspondence between the two classes. So we have proved a theorem, which is a wrong one! because we have not taken account of the exceptional case of pentagonal numbers. The fallacy lies in having overlooked the fact that the last slope may extend right up to the ﬁrst row; the slope and the row may very well interfere. Let us take one such instance. Let again s < r.

If we place the last slope above the ﬁrst row this works because the number 31 of pointsin the ﬁrst row is also diminished by one, in fact by the disputed point (notice again that no two rows are equal for s < r 1). So the interference is of no account. With s r we may again have an interfering− case. We again place the top row ≥ 4. Lecture 26

behind the last slope, this time with a punishment. We have now shortened the slope by 1. For s 1 r the method is still good. So the only cases of earnest interference are: − ≥

(i) s < r but x r 1. Then r 1 s r and hence s = r 1 ≥ − − ≤ ≤ − (ii) s r but s 1 < r. Then s r > s 1 and hence s = r. ≥ − ≥ − Here we have something which can no longer be overcome. These are the cases of pentagonal numbers. In (ii) the total number of dots is equal to

s + (s + 1) + (s + 2) + + (2s 1) ··· − s(3s 1) = − 2 In (i) this number = (s + 1) + (s 2) + + 2s + − ··· = s(3s 1) 2 These decompositions do not have companions. In general every partition into one parity of diﬀerent summands has a companion of the other parity of diﬀerent summands; and in the case of pentagonal numbers there is just one in 32 excess in one of the classes. We now come to the most important application of identity (1). Since

1 = ∞ n m p(n)x , m∞=1(1 x ) = − Xn 0 we have on combining thisQ with (1),

∞ n ∞ λ λ(3λ 1)/2 1 = p(n)x ( ) x − (2) − n=0 λ= X X−∞ 4. Lecture 27

This tells us the follwing story. All the coefficients on the right side of (2) excepting the first must be zero. The typical exponent in the second factor on = the right side is λ(3λ 1)/2 ωλ, say. (The first few ωλ′ s are 0, 1, 2, 5, 7, 12, 15, ...). Now look for− xn. Since the coefficient in the first factor is p(n) and that in the second always 1, we have, since xn(n , 0) does not appear on the left side ±

p(n) p(n 1) p(n 2) + p(n 5) + p(n 7) ++ = 0 − − − − − − −− ··· or

λ p(n ωλ)( ) = 0 (3) − − 0 ω n ≤Xλ≤ This is a formula of recursion. Omitting the ﬁrst index of summation (3) gives λ 1 p(n) = ( ) − p(n ωλ) (4) − − 0<ω n Xλ≤ Let us calculate the ﬁrst few p(n).

p(0) = 1 p(1) = p(1 1) = p(0) = 1 − p(2) = p(2 1) + p(2 2) = 2 − − p(3) = p(3 1) + p(3 2) = 3 − − p(4) = p(4 1) + p(4 2) = 5 − − p(5) = p(5 1) + p(5 2) p(5 5) = 7 − − − −

(Watch! a pentagonal number - and a negative sign comes into action!). These 33 formulae get longer and longer, but not excessively so. Let us estimate how λ(3λ 1) long these will be. Since ωλ n we have to look for λ satisfying − ≤ 2 ≤ n, which gives

12λ(3λ 1) 24n, − ≤ 36λ2 12λ 24n, − ≤ (5λ 1)2 = 24n + 1, − 6λ 1 = √24n + 1, | − | 4. Lecture 28

1 1 λ √24n + 1. | − 6|≤ 6 1 2 Hence roughly there will be √24n = √6n summands on the left side 3 3 of (3). So their number increases with the square root of n- the expressions do not get too long after all (for n = 100, we have 17 terms). These formulae have been used for preparing tables of p(n) which have been quite useful. For instance Ramanujan discovered some of the divisibility properties of p(n) by using them. In the famous paper of Hardy and Ramanu- 34 jan (1917) there is a table of p(n) for n 200. These were computed by Macmahon, by using the above formulae≤ and the values were checked with those given by the Hardy-Ramanujan formula. The asymptotic values were found to be very close to what Macmahon computed. Gupta has extended the table for p(n) up to 600. Before making another application of Euler’s pentagonal theorem, we proceed a bit further into the theory of formal power series. We add now one more formal procedure, that of formal diﬀerentiation. Let

2 A = a + a1 x + a2x + ◦ ···

The derivative A′ of A is by deﬁnition

2 A′ = a + 2a x + 3a x + 1 2 3 ··· This is again a power series in our sense. This operation of diﬀerentiation which produces one power series from another is a linear operation:

(A + B)′ = A′ + B′, where B is a second power series. This is easy to verify; actually we need do this only for polynomials as everything is true modulo xN . Again,

(cA)′ = cA′ as can be seen directly. Also

(A B)′ = A′B + AB′. · Let us look into this situation. Start with the simplest case, A = xm, B = xn. Then

m 1 n 1 A′ = mx − , B′ = nx − 4. Lecture 29

m+n m+n 1 and (AB)′ = (x )′ = (m + n)x − , m 1+n m+n 1 also A′ B + AB′ = mx − + nx − m+n 1 = (m + n)x −

So this is true also for polynomials by linearity, we can do it piecemeal. 35 And as it is enough if we stop short at xN , it is true in general, Let us add one more remark. Let us write down a special case where A and B have reciprocals. Then AB has a reciprocal too (since the units form a group). In this case we have

(AB) A B ′ = ′ + ′ , AB A B which is the rule for logarithmic diﬀerentiation. (It is identical with the procedure in the calculus, as soon as we speak of functions). For A, B and C,

(ABC)′ = A′(BC) + A(BC)′ = A′BC + AB′C + ABC′ (ABC) A B C or ′ = ′ + ′ + ′ , ABC A B C and so on; in general, K K = A ′ n 1 k = A′K K A Q k=1 Ak k=1 k X We can do this for inﬁniteQ products also if the products are permissible. K ℓ N Indeed Ak is legitimate if Aℓ = 1 + aℓ(k) x + Consider modulo x ; break k=1 ··· at a ﬁniteQ spot and the factors 1 will come into action. Lecture 5

Let us consider some applications of formal diﬀerentiation of power series. 36 Once again we start from the pentagonal numbers theorem:

∞ m ∞ λ λ(3λ 1)/2 (1 x ) = ( ) x − − − m=1 λ= Y X−∞ ∞ = ( )λxωλ , (1) λ= − X−∞ λ(3λ 1) with ω = − . Taking the logarithmic derivative - and this can be done λ 2 piecemeal-

∞ λ ωλ 1 m 1 ( ) ωλ x − ∞ mx − λ= − − = −∞ 1 xm P m=1 − ∞ ( )λxωλ X λ= − −∞ Multiplying both sides by x, P

∞ λ ωλ m ( ) ωλ x ∞ mx λ= − = −∞ − m (2) 1 x P∞ m=1 − ( )λxωλ X λ= − −∞ The left side here is an interesting objectP called a Lambert series, with a structure not quite well deﬁned; but it plays some role in number theory. Let us transform the Lambert series into a power series; it becomes

∞ ∞ ∞ m xkm = mxkm, − = = − = Xm 1 Xk 1 XXk1m 1 30 5. Lecture 31

and these are all permissible power series, because though there are inﬁnitely many of them, the inner ones begin with later and later terms. Rearranging, this gives 37

∞ ∞ mxn = xn m − = = − = nXkm Xm 1 Xn 1 Xm/n ∞ = σ(n)xn, − = Xn 1 where σ(n) denotes the sum of the divisors of n, σ(n) = d. d n | (Let us study σ(n) for a moment. P

σ(1) = 1, σ(2) = 3, σ3 = 4, σ(5) = 6; indeed σ(p) = p + 1 for a prime p. And σ(n) = n + 1 implies that n is prime. σ(n) is not too big; there can be at most n divisers of n and so roughly σ(n) = O(n2). In fact it is known that σ(n) = O(n1+epsilon), > 0, that is, a little larger than the ﬁrst power. We shall however not be studying∈ σ(n) in detail). Equation (2) can now be rewritten as

∞ ∞ ∞ n λ ωλ λ 1 ωλ σ(n)x ( ) x = ( ) − ωλ x − − n=1 λ= λ= X X−∞ X−∞ Let us look for the coefficient of xm on both sides. Remembering that the m first few ω′ s are 0, 1,2, 5,7, 12,15 , the coefficient of x on the left side is λ ··· σ(m) σ(m 1) σ(m 2) + σ(m 6) + σ(m 7) ++ − − − − − − −− ··· On the right side the coefficient is 0 most frequently, because the pentago- λ 1 nal numbersare rather rare, and equal to ( ) ωλ exceptionally, when m = ωλ. − −

++ = 0 usually, σ(m) σ(m 1) σ(m 2) λ 1 − − − − −−··· ( ) ωλ for m = ωλ.  − −  We now single out σ(m).  38 We may write

, = λ 1 + 0 usually σ(m) ( ) − σ(m ωλ) λ 1 − − ( ) − ωλ for m = ωλ 0<ωλ

This is an additive recursion formula for σ(n). We can make it even more striking. The inhomogeneouspiece on the right side is a little annoying. σ(m − m) can occur on the right side only for m = ωλ; σ(0) does not make sense; however, for our purpose let us deﬁne

σ(m m) = m. −

Then σ(ωµ ωµ) = ωµ, and the previous formula can now be written uninterruptedly as− λ 1 σ(m) = ( ) − σ(m ωλ) (3) − − 0<ω m Xλ≤ We have proved earlier that

λ 1 p(m) = ( ) − p(m ωλ) (4) − − 0<ω m Xλ≤ which is a formula completely identical with (3). Here p(m m) = p(0) = 1. It is extraordinary that σ(m) and p(m) should have the same recursion− formula, diﬀering only in the deﬁnition of the term with n = 0. This fact was noted by Euler. In fact p(m) is increasing monotonically, while the growth of σ(m) is more erratic. There are more relations between p(m) and σ(m). Let us start again with the identity ∞ ∞ (1 xm) p(m)xm = 1 (5) = − = Ym 1 Xm 0 We know that for a pair of power series A, B such that AB = 1, on taking 39 A B A B logarithmic derivatives, we have ′ + ′ = 0 or ′ = ′ . So from (5), A B A − B

∞ np(n)xn ∞ = σ(m)xm = n 0 , P m=1 ∞ p(n)xn X n=0

∞ ∞ P ∞ or σ(m)xm p(k)xk = np(n)xn. = = = Xm 1 Xk 0 Xn 0 Comparing coeﬃcients of xn,

np(n) = σ(m)p(k), + = mXk n 5. Lecture 33

or more explicitly, ∞ np(n) = σ(m)p(n m) (6) = − Xm 1 This is a bilinear relation between σ(n) and p(n). This can be proved directly also in the following way. Let us consider all the partitions of n; there are p(n) such:

n = h + h + 1 2 ··· n = k + k + 1 2 ··· n = ℓ + ℓ + 1 2 ··· ......

Adding up, the left side gives np(n). Let us now evaluate the sum of the right sides. Consider a particular summand h and let us look for those partitions in which h ﬁgures. These are p(n h) partitions in which h occurs at least once, p(n 2h) in which h occurs at least− twices; in general, p(n rh) in which h occurs− at least r times. Hence the number of those partitions− which contain h exactly r times is p(n nh) p(n n + 1h). Thus the number of times h occurs 40 in all partitions put together− − is −

n p(n nh) (n n + 1h) − − − nh n X≤ n o Hence the contribution from these to the right side will be

h n p(n nh) (n n + 1h) = h p(n nh) − − − − nh n nh n X≤ n o X≤ on applying partial summation. Now summing over all summands h, the right side becomces m h p(n nh) = p(n m), − n − h nh n n/m m n X X≤ X X≤ on putting rh = m; and this is

m n p(n m) = p(n m)σ(m). − n − m n n.m m=1 X≤ X X Let us make one ﬁnal remark. 5. Lecture 34

Again from the Euler formula,

∞ λ 1 ωλ ( ) − ωλ x ∞ m λ= − σ(m)x = −∞ P m=1 ∞ ( )λxωλ X λ= − −∞ ∞ P λ 1 ω ( ) − ωλ x λ λ= − = −∞ P ∞ (1 xm) m=1 −

∞ Q ∞ λ 1 ωλ m = ( ) − ωλ x p(m)x − λ= m=0 X−∞ X Comparing the coeﬃcients of xm on both sides, 41

σ(m) = p(m) 1 p(m 1) 2 p(m 2) + 5 p(m 5) − · − − · − · − + 7 p(m 7) + · − − ··· λ 1 = ( ) − ωλ p(m ωλ) − − 0 ω m ≤Xλ≤ This last formula enables us to ﬁnd out the sum of the divisors provided that we know the partitions. This is not just a curiosity; it provides a useful check on tables of pertitions computed by other means. We go back to power series leading up to some of Ramanujan’s theorems. Jacobi introduced the products

∞ 2n 2n 1 1 2n 1 (1 x )(1 + zx − )(1 + z− x − ). = − Yn 1 This is a power series in x; though these are infinitely many factors they start with progressively higher powers. The coefficients this time are not polynomials in z but from the field R(z), the field of rational functions of z, which is a perfectly good field. Let us multiply out and we shall have a very nice surprise. The successive coeffieicnts are: 1 1 1 x : z + z− (note that this is unchanged when z z− ) x2 : (1 + 1) = 0 → 3 1 1 x :(z + z− z z− ) = 0 4 − 2− 2 2 2 1 x :( 1 1 + z + 1 + 1 + z− ) = z + z− (again unchanged when z z− ) ...... − − → 5. Lecture 35

We observe that non-zero coeﬃeicnts are associated only with square ex- 42 ponents. We may threfore provisionally write

∞ 2n 2n 1 1 2n 1 ∞ k k k2 (1 x )(1 + zx − )(1 + z− x − ) = 1 + (z + z− )x = − = Yn 1 Xk 1 ∞ 2 = zk xk (7) k= X−∞ (with the terms corresponding to k folder together). This is a - series; only quadratic exponents occur. ± V We shall nowprovetheidentity(7). But wehavegotto becareful. Consider the polynomial

N 2n 2n 1 1 2n 1 ΦN (x, z) = (1 x )(1 + zx − )(1 + z− x − ) = − Yn 1 This consists of terms z j xk with N j N,0 k N(N + 1) + 2N2 = 3N2 + N. We can rearrange with respect− ≤ to≤ powers≤ of z.≤ The coeﬃcients are 1 now polynomials in x. z and z− occur symmetrically.

1 2 2 N N ΦN(x, z) = C (x) + (z + z− )C1(x) + (z + z− )C2(x) + + (z + z− )CN (x). ◦ ···

Let us calculate the C′ s. It is cumbersome to look for C , for so many ◦ cancellations may occur. It is easier to calculate CN . Since the highest power of z can occur only from the terms with the highest power of x, we have

N 2n 1+3+ +(2N 1) CN (x) = (1 x ) x ··· − = − × Yn 1 N 2 = xN (1 x2n) = − Yn 1 2 Now try to get a recursion among the C′ s. Replacing z by zx , we get 43

N 2 2n 2n+1 1 2n 3 ΦN(x, zx ) = (1 x )(1 + zx )(1 + z− x − ). = − Yn 1 2 Compare ΦN(x, zx ) and ΦN(x, z); these are related by the equation

2 1 2N 1 2n+1 1 1 ΦN(x, zx )(1 + zx)(1 + z− x − ) =ΦN (x, z)(1 + zx )(1 + z− x− ) 5. Lecture 36

The negative power in the last factor on the right is particularly disgusting; to get rid of it we multiply both sides by xz, leading to

2 2N 2N+1 ΦN (x, zx )(xz + x ) =ΦN(x, z)(1 + zx ), 2N+1 1 N N or (1 + zx )(C (x) + (z + z− )C1(x) + + (z + z− )CN (x)) ◦ ··· 2n 2 1 2 = (xz + x )(C (x) + (zx + z− x− )C1(x)+ ◦ 2 4 2 4 N 2N N 2N + (z x + z− x− )C (x) + + (z x + z− x− )C (x)) 2 ··· N These are perfectly harmless polynomials in x; we may compare coeﬃ- cients of zk. Then

2N+1 2k+2N 2k 1 Ck(x) + Ck 1(x)x = Ck(x)x + x − Ck 1(x), − − 2N+2k 2k 1 2N 2k+2 or Ck(x)(1 x ) = Ck 1(x)x − (1 x − ) − − − (We proceed from Ck to Ck 1 since CN is already known). − 2k+1 2N+2k = x− (1 x ) Ck 1(x) − + Ck(x) − 1 x2N 2k 2 − − N N2 2n Since CN (x) = x (1 x ), we have in succession 44 n=1 − Q 4N N = N2 2N+1 1 x 2n CN 1(x) x − − 2 (1 x ) − 1 x = − − Yn 1 N (N 1)2 2n 4N = x − (1 x ) (1 x ); = − · − Yn 2 N (N 2)2 2n 4N 4N 2 CN 2(x) = x − (1 x ) (1 x )(1 x − ) − = − · − − Yn 3 ......

In general,

N j 1 − (N j)2 2n 4N 2m CN j(x) = x − (1 x ) (1 x − ) − = + − = − nYj 1 Ym 0 or, with j = N n, − N N n 1 n2 2n − − 4N 2m C (x) = x (1 x ) (1 x − ) (8) n − − n=N n+1 m=0 Y− Y 5. Lecture 37

Equation (8) leads to some congruence relations. The lowest terms of Cn(x) have exponent

n2 + 2(N n + 1) = 2N + (n2 2n + 1) + 1 2N + 1 − − ≥ Hence 2 C (x) xk (mod x2N+1) (9) n ≡ From the original formula,

N 2n 2n+1 1 2n 1 Φ (x, z) = (1 x )(1 + zx )(1 + z− x − ) N − n=1 Y 1 2 2 4 2N+1 1 + (z + z− )x + (z + z− )x + (mod x ) ≡ ··· ∞ 2 zk xk (mod x2N+1), ≡ k= X−∞ since the inﬁnite series does not matter, the higher powers being absorbed in 45 the congruence. Hence

∞ 2n 2n 1 1 2n 1 2N+1 ΦN (x, z) (1 x )(1 + zx − )(1 + z− x − ) (mod x ) ≡ = − Yn 1 The new terms x2N+2,..., areabsorbedby mod x2N+1. We have

∞ 2n 2n 1 1 2n 1 ∞ k k2 2N+1 (1 x )(1 + zx − )(1 + z− x − ) z x (mod x ) − ≡ n=1 k= Y X−∞ Thus both expansions agree as far as we wish, and this is what we mean by equality of formal power series. Hence we can replace the congruence by equality, and Jacobi’s identity (7) is proved. As an application of this identity, we shall now give a new proof of the pentagonal numbers theorem. We replace x by y3, as we could consistently in the whole story; only read modulo y6N+3. Then we have

∞ 6n 6n 3 1 6n 3 ∞ k 3k2 (1 y )(1 + zy − )(1 + z− y − ) = z y − n=1 k= Y X−∞ We now do something which needs some justiﬁcation. Replace z by y. This is something completely strange, and would interfere seriously with− our 3 6N+3 reasoning. For ΦN (y , z) we had congruences modulo y . If we replaced z 5. Lecture 38

by y3 nobody could forbid that. Since z occurs in negative powers, the powers of y might be lowered too by as much as N. We obtain polynomials in y alone + on both sides, but true modulo y5N 3, because we may have lowered powers of 46 y. With this proviso it is justiﬁed to replace z by y; so that ultimately we have −

∞ 6 6n 2 6n 4 ∞ k 3k2+k 5N+3 (1 y )(1 y − )(1 y − ) = ( ) y (mod y ) − − − − n=1 k= Y X−∞ We can carry over the old proof step by step. Since we now have only even powers of y, this leads to

∞ ∞ (1 y2m) = ( )kyk(3k+1) − − m=1 k= Y X−∞ These are actually power series in y2. Set y2 = x, then

∞ ∞ (1 xm) = ( )K xk(3k+1)/2 − − m=1 k= Y X−∞ which is the pentagonal numbers theorem. Lecture 6

In the last lecture we used the Jacobi formula: 47

∞ 2n 2n 1 1 2n 1 ∞ k k2 (1 x )(1 + zx − )(1 + z− x − ) = z x (1) − n=1 k= Y X−∞ to give a new proof of Euler’s pentagonal numbers theorem. We proceed to give another application. We observe again that the right side of (1) is a power series in x; we cannot do anything about the z′s and no formal diﬀerentiation can be carried out with respect to z. Let us make the substitution z zx. This again interferes greatly with our variable x. Are we entitled to do→− this? Let us look back into our proof of (1). We started with a curtailed aﬀair

∞ 2n 2n 1 1 2n 1 ΦN (x, z) = (1 x )(1 + zx − )(1 + z− x − ) = − Yn 1 and this was a polynomial of the proper size and everything went through. When we replace z by zx and multiply out, the negative powers might accumulate and we might be− destroying xN possibly; nevertheless the congruence relations would be true this time modulo xN+1 instead of x2N+1 as it was previously; but this is all we went. So the old proof can be reproduced step by step and every thing matches modulo xN+1. (Let us add a side remark. In the proof of (1) we had to replace z by zx2 - and this was the essential step in the proof. We cannot do the same here as this would lead to congruences mod x only. Before we had the congruences we had identities and there we could carry out any substitution. Then we adopted a new point of view and introduced congru- 48 ences; and that step bars later the substitution z zx2. So let us make the substitution z zx without→ further compuncton. This →−

39 6. Lecture 40

gives us

∞ 2n 2n 1 2n 2 ∞ k k k2+k (1 x )(1 zx )(1 z− x − ) = ( ) z x − − − − n=1 k= Y X−∞ This is not nicely arranged. There appears an extraordinary term without x- corresponding to n = 1 in the last factor on the left side; let us keep this apart. Also on the right side the exponentof x is k(k +1), so that every number occurs twice; let us keep these two pieces together. We then have

1 ∞ 2n 2n 1 2n (1 z)− (1 x )(1 zx )(1 z− x ) − = − − − Yn 1 ∞ k k k(k+1) ∞ k 1 k 1 k(k+1) = ( ) z x + ( )− − z− − x = − = − Xk 0 Xk 0 (where in the second half we have replaced k by k 1), − − ∞ k k(k+1) k k 1 = ( ) x (z z− − ) = − − Xk 0 ∞ k k(k+1) k 2k 1 = ( ) x z (1 z− − ) = − − Xk 0 ∞ k k(k+1) k 1 1 2 2k = ( ) x z (1 z− )(1 + z− + z− + + z− ) = − − ··· Xk 0 We now have an infinite series in x equal to another. Now recollect that our coefficients are from the field R(z) which has no zero divisors. So we may 1 cancel 1 z− on both sides; this is a non-zero factor in R(z) and has nothing to 49 do with di−fferentiation. This leads to

∞ 2n 2n 1 2n ∞ k k(k+1) k k 1 k (1 x )(1 zx )(1 z− x ) = ( ) x (z + z − + + z− ). = − − − = − ··· Yn 1 Xk 0 In the ﬁeld R(z) we can replace z by 1. We can do what we like in the ﬁeld and that is the essence of the power series method. So putting z = 1,

∞ ∞ (1 x2n)3 = ( )k xk(k+1)(2k + 1). = − = − Yn 1 Xk 0 This is a power series in x2; give it a new name, x2 = y. Then

∞ ∞ (1 yn)3 = ( )k(2k + 1)yk(k+1)/2 (2) = − = − Yn 1 Xk 0 6. Lecture 41

This is a very famous identity of Jacobi, originally proved by him by an altogether diﬀerent method using the theory of functions. Let us juxtapose it with the Euler pentagonal formula:

∞ n ∞ λ λ(3λ 1)/2 (1 y ) = ( ) x − (2a) − − n=1 λ= Y X−∞ Let us proceed to yet another application of the triple product formula; we shall obtain some of Ramanujan’s formulas. Taking away the ﬁrst part of the triple product formula we have

∞ 2n 1 1 2n 1 ∞ k k2 1 (1 + zx − )(1 + z− x − ) = z x (3) n=1 k= ∞ (1 x2n) −∞ Y X n=1 − Q The second part on the right side here is of interest, because it is the gener- 50 ating function of the partition. We had earlier the formula

∞ 2n 1 ∞ m (1 + zx − ) = z Cm(n), = = Yn 1 Xm 0 (4) xm2 C (x) = m (1 x2) (1 x2m) − ··· − and these are permissible power series, beginning with later and later powers of x, and so the right side of (4) makes sense, as a formal power series in x. Substituting (4) in (3), we have

∞ n ∞ s ∞ k k2 1 z Cr(x) z− Cs(x) = z x (5) r=0 s=0 k= ∞ (1 x2n) −∞ X X X n=1 − Q We can compare zO on both sides for, for very high xN the left side will contain only ﬁnitely many terms and all otheres will disappear below the hori- zon; we can also add as many terms as we wish. So equating coeﬃcients of zO, we have

∞ 1 Cr(x)Cr(x) = , r=0 ∞ (1 x2n) X n=1 − 2 ∞ x2r Q 1 or = (1 x2)2 (1 x2n)2 r=0 − ··· − ∞ (1 x2n) X n=1 − Q 6. Lecture 42

We have even powers of x consistently on both sides; so replace x2 by y, and write down the ﬁrst few terms explicitly:

y y4 y9 1 + + + + (1 y)2 (1 y)2(1 y2)2 (1 y)2(1 y2)2(1 y3)2 ··· − − − − − − 1 = (6) ∞ (1 yn) n=1 − Q This formula is found in the famous paper of Hardy and Ramanujan (1917) 51 and ascribed by them to Euler. It is very useful for rough appraisal of asymptotic formulas. Hardy and Ramanujan make the cryptic remark that it is “a formula which lends itself to wide generalisations”. This remark was at ﬁrst not very obvious to me; but it can now be interpreted in the following way. Let us look for zk in (5). Then

xk2 Cr(x)Cs(x) = r,s ∞ (1 x2n) rXs=k − n=1 − Q or, replacing r by s + k, and writting Cs for Cs(x), the left side becomes

2 + + 2 ∞ xk x1 (k 1) + = + + CsCs k 1 2 2k 2 2 4 2k+2 = · (1 x ) (1 x ) (1 x ) (1 x ) (1 x ) Xs 0 − ··· − − − ··· − 4+(k+2)2 + x + (1 x2)2(1 x4)2(1 x6) (1 x2k+4) ··· − − − ··· − 2 Let us divide by xk . The general exponent on the right side is ℓ2 + (k + ℓ)2, 52 so on division it becomes 2ℓ2 + 2kℓ. Every exponent is even, which is a very nice situation. Replace x2 by y, and we get the ‘wide generalisation’ of which Hardy and Ramanujan spoke:

k+1 1 + y (1 y)(1 y2) (1 yk) (1 y)2(1 y2) (1 yk+1) − − ··· − −2(k+2) − ··· − + y + (1 y)2(1 y2)2(1 y3) (1 yk+2) ··· − − − ··· − yl(k+l) 1 + = (7) (1 u)2 (1 yl)2(1 yl+1) (1 yk+l) ··· − ··· − − ··· − ∞ (1 yn) n=1 − Q 6. Lecture 43

k is an assigned number and it can be taken arbitrarily. So such expansions are not unique. Thus (6) and (7) give two different expansions for 1 . ∞ (1 yn) n=1 − Q We are now slowly coming to the close of our preoccupation with power series; we shall give one more application due to Ramanujan (1917). In their paper Hardy and Ramanujan gave a surprising asymptotic formula for p(n). 1/4 It contained an error term which was something unheard of before, (n− ), error term decreasing as n increases. Since p(n)is anintegerit isenoughtoO take a few terms to get a suitable value. The values calculated on the basis of the 53 asymptotic formula were checked up with those given by Macmahon’s tables and were found to be astonishingly close. Ramanujan looked at the tables and with his peculiar insight discovered something which nobody else could have noticed. He found that the numbers p(4), p(9), p(14), in general p(5k + 4) are all divisible by 5; p(5), p(12), p(7k + 5) are all divisible by 7; p(11k + 6) by 11. So he thought thiswas a general··· property. A divisibility property of p(n) is itself surprising, because p(n) is a function defined with reference to addition. The first and second of these results are simpler than the third. Ramanujan in fact suggested more. If we chose a special progression modulo5λ, then all the λ 2λ 1 terms are divisible by 5 . There are also special progressions modulo 7 − ; so for 11. Ramanujan made the general conjecturethat if δ = 5a7b11c and 24n 1 (mod δ), then p(n) 0 (mod δ). In this form the conjecture is wrong. These≡ things are deeply connected≡ with the theory of modular forms; the cases 5 and 7 relate to modular forms with subgroups of genus 1, the case 11 with genus 2. Let us take the case of 5. Take p(5k + 4). Consider Σp(n)xn; it is nicer to multiply by x and look for x5k. We have to show that the coefficients of x5k in xΣp(n)xn are congruent to zerp modulo 5. We wish to juggle around n 5k with series a bit. Take Σan x ; we want to study x . Multiply by the series 1 + b x5 + b x10 + where the b s are integers. We get a new power series 1 2 ··· ′ a xn (1 + b x5 + b x10 + ) = c xn, n · 1 2 ··· n which is just asX good. It is enough if we prove thatX for this series every fifth 54 coefficient 0 (mod 5). For, ≡

n n = cn x anx 5 10 1 + b1 x + b2 x + X P ··· 6. Lecture 44

= c xn, (1 + d x5 + d x10 + ), say. n 1 2 ··· n Then if every fifth coeXfficient of Σcn x is divisible by 5, multiplication by 5n Σdn x will not disturb this. For a prime p look at p p p p (1 + x)p = 1 + x + x2 + x3 + + xp. 1 2 3 ··· p ! ! ! ! All except the first and last coefficients on the right side are divisible by p, p = p! for in a typical term q (p q)!q! , the p inm the numerator can be cancelled only by a p in the denominator.− So (1 + x)p 1 + xp (mod p). ≡ This means that the difference of the two sides contains only coefficients divisible by p. This (1 x)5 1 + x5 (mod 5) − ≡ We now go to Ramanujan’s proof that p(5k + 4) 0 (mod 5) We have 55 ≡ x x p(n)xn = (1 xn) X − It is irrelevant here if we multiply bothQ sides by a series containing only x5, x10, x15, . This will not ruin our plans as we have declared in advance. So ···

n ∞ 5m = x ∞ 5m x p(n)x (1 x ) n (1 x ) = − (1 x ) = − X Ym 1 − Ym 1 Q x ∞ m 5 n (1 x ) modulo5 ≡ (1 x ) = − − Ym 1 (1 x5m) (1 xm)5hasQ only coeﬃcients divisible by 5 − − − Y Y ∞ x (1 xm)4 modulo 5 ≡ = − Ym 1 ∞ ∞ = x (1 xm) (1 xm)3. = − = − Ym 1 Ym 1 For both products on the right side we have available wonderful expressions. By (2) and (2a),

m m 3 ∞ λ(3λ 1)/2 ∞ k k(k+1)/2 x (1 x ) (1 x ) = x ( ) − ( ) (2k + 1)x − − − − λ= k=0 Y Y X−∞ X 6. Lecture 45

The typical term on the right side is

∞ λ+k 1+λ(3λ 1)/2 ( ) x − + k(k + 1)/2 = − Xk 0 The exponent = 1 + λ(3λ 1)/2 + k(k + 1)/2, and we want this to be of the − form 5m. Each such combination contributes to x5m. We want 56 λ(3λ 1) k(k + 1) 1 + − + 0 (mod5) 2 2 ≡ Multiply by 8; that will not disturb it. So we want

8 + 12λ2 4λ + 4k2 + 4k 0(5), − ≡ 3 + 2λ2 4λ + 4k2 + 4k 0(5), − ≡ 2(λ 1)2 + (2k + 1)2 0(5). − ≡ This is of the form: 2. a square + another square 0(5) ≡ Now

A2 0, 1, 4(5), ≡ 2B2 0, 2, 3(5); ≡ and so A2 + 2B2 0(5) means only the combination A2 0(5) and 2B2 0(5); each square must≡ therefore separately be divisible by 5,≡ or ≡

2k + 1 0(5) ≡ So to x5m has contributed only those combinationsin which 2k+1 appeared; andeveryoneof these piecescarriedwith it a factor of5. This porvesthe result. The case 7k + 5 is even simpler. We multiply by a series in x7 leading to (1 xm)6 which is to be broken up into two Jacobi factors (1 xm)3. These are examples− of very beautiful theorems proved in a purely forma−l way. We shall deal in the next lecture with one more starting instance, the Rogers-Ramanujan identities which one cannot refrain from talling about. Lecture 7

We wish to say something about the celebrated Rogers-Ramanujan identities: 57

x x4 x9 1 + + + 1 x (1 x)(1 x2) (1 x)(1 x2)(1 x3) − − − − − − 1 + = ; (1) ··· n>0 (1 xn) n 1Q (mod 5) ≡± − x2 x2 3 x3 4 1 + + · + · 1 x (1 x)(1 x2) (1 x)(1 x2)(1 x3) − − − − − − 1 + = (2) ··· n>0 (1 xn) n 2Q (mod 5) ≡± − The right hand sides of (1) and (2), written down explicitly, are respectively 1 (1 x)(1 x4)(1 x6)(1 x9) ... − − − − 1 (1 x2)(1 x3)(1 x7)(1 x8) ... − − − − One immediately observes that 1 are quadratic residues modulo 5, and 2 quadratic non-residues modulo 5.± These identities were ﬁrst communicated ±by Ramanujan in a letter written to Hardy from India in February 1913 before he embarked for England. No proofs were given at that time. It was a remarkable fact, nevertleless, to have even written down such identities. It is 58 true that Euler himself did some experimental work with the pentagonal numbers formula. But one does not see the slightest reason why anybody should have tried 1, 2 modulo 5. Then in 1917 something happened. In an old ± ±

46 7. Lecture 47

volume of the Proceedings of the London Mathematical Society Ramanujan found that Rogers (1894) had these identities along with extensions of hypergeometric functions and a wealth of other formulae. In 1916 the identities were published in Macmahon’s Combinatory Analysis without proof, but with a number-theoretic explanation. This was some progress. In 1917 I.Schur gave proofs, one of them combinatorial, on the lines of F.Franklin’s proofof Euler’s theorem. Schue also emphasized the mathematical meaninf of the identities. Let us look at the meaning of these identities. Let us write the right side of (1) as a power-series, say,

1 ∞ = n, 4 6 9 q′(n)x (1 x)(1 x )(1 x )(1 x ) ... = − − − − Xn 0 q′(n) is the number of terms collected from summands 1, 4, 6, ... with repetitions, or, what is the same thing, the number of times in which n can be expressed as the sum of parts 1 (mod 5), with repetitions. Likewise, if we write ≡± 1 ∞ n = q′′(n)x , (1 xn) n=0 n 2(5) − ≡± X Q then q′′(n) is the number of representations of n as the sum of parts 2 (mod 5), with repetitions. ≡ ± The expressions on the other side appear directly. Take 59 xk2 (1 x)(1 x2) (1 x4) − − ··· − If we write 1 = a + a x + a x2 + (1 x)(1 x2) (1 xk) 0 1 2 ··· − − ··· − then the coeﬃcient an gives us the number of partitions of n into parts not exceeding k. Let us represent the partitions by dots in a diagram, each vertical column denoting a summand. Then there are at most k rows in the diagram. Since k2 is the sum of the k ﬁrst odd numbers,

k2 = 1 + 3 + 5 + + (2k 1), ··· − each partition of n into summands not exveeding k can be enlarged into a partition of n + k2 into summands which diﬀer by at least two, for we can adjoin k2 dots on the left side, putting one in the lowest row, three in the next, ﬁve 7. Lecture 48

in the one above and so on ﬁnally 2k 1 in the top most row. Conversely any partition of n into −

parts with minimal difference 2 can be mutilated into a partition of n k2 into 60 summands not exceeding k. Hence there is a one one correspondence− between these two types. So the coefficients in the expansion of xk2 represent the number of times that a number N can (1 x)(1 x2) (1 xk) be− decomposed− ··· into k−parts (the partitions are now read horizontally in the diagram) differing by two at least. When this is done for each k and the results added up, we get the following arithmetical interpretation of (1): The number of partitions of n with minimal difference two is equal to the number of partitions into summands congruent to 1 (mod 5) allowing repetitions. A similar explanation is possible in± the case of (2). On the left side we can account for the exponents 2.3, 3.4, ..., k(k + 1), ... in the numerator by means of triangular numbers. In the earlier diagram we adjoin on the left 2, 4, 6, ..., 2k dots beginning with the lowerst row. The number thus added is 2+4+ +2k = k(k+1); this disposes of xk(k+1)in the numerator. So read horizontally,········· the diagram gives us a decomposition into parts which differ by xk(k+1) 2 at least, but the summand 1 is no longer tolerated. gives (1 x) (1 xk) us therefore the enumeration of xN by parts differeing by− 2 at··· least,− the part 1 being forbidden. We have in this way the following arithmetical interpretation of (2): The number of partitions of n into parts not less than 2 and with minimal difference 2, is equal to the number of partitions of n into parts congruent 2 (mod 5), repetitions allowed. ± By a similar procedure we can construct partitions where 1 and 2 are for- 61 bidden, partitions differing by at least three, etc. In the case where the difference is 3, we use 1, 4, 7,..., so that the number of dots adjoined on the left is 1 + 4 + 7 + to k terms = k(3k 1)/2, so a pentagonal number, and this is ··· − 7. Lecture 49

xk(3k 1)/2 no surprise. In fact − would give us the number of (1 x)(1 x2) (1 xk) ff − − ··· − partitions into partsP di ering by at least 3. And for 4 the story is similar. The unexpected element in all these cases is the association of partitions of a definite type with divisibility properties. The left-side in the identities is trivial. The deeper part is the right side. It can be shown that there can be no corresponding identities for moduli higher than 5. All these appear as wide generalisations of the old Euler theorem in which the minimal difference between the summandsis, of course, 1. Euler’s theoremis thereforethe nucleus of all such results. We give here a proof of the Roger-Ramanujan identities which is in line with the treatment we have been following, the motiod of formal power series. It is a transcription of Roger’s proof in Hardy’s ‘Ramanujan’, pp.95-98. We use the so-called Gaussian polynomials. Let us introduce the Gaussian polynomials in a much neater notation than usual. Consider for first the binomial coefficients: n n(n 1)(n 2) (n k + 1) = − − ··· − m 1 2 3 k ! · · ···· (Observe that both in the numerator and in the denominator there are k fac- 62 tors, which are consecutive integers, and that the factors of equal rank in both + n numerator and denominator always add up to n 1). The k are all integers, as is obvious from the recursion formula + n 1 = n + n k k k 1 ! ! − ! n = n = n = n 1, of course, and by definition, 0 1 We also define k 0 for k > n n = n or for k < 0. Observe also the eymmetry: k n k The Gaussian polynomials are something of a− similar nature. We define the Gaussian polynomial

n = n k k " # " #x n (1 xn)(1 xn 1) (1 xn k+1) by = − − − · − − k (1 x)(1 x2) (1 xk) " # − − ··· − The sum of the indices of x in corresponding factors in the numerator r and n denominator is n + 1, as in n . That the are polynomials in x is obvious k k " # 7. Lecture 50

from the recursion formula n + 1 n n + 1 = + xk k k k 1 " # " # " − # n n where = 1 and = 1 by deﬁnition. The recursion formula is just the same n 0 " # n " # as that for k except for the factor in the second term on the right. Also deﬁne 0 n = 1; also let = 1 for k > n or k < 0. 63 0 0 " # " # 1 0 0 = + x = 1, 0 0 1 ◦ " # " # "− # 1 1 = ; 1 0 " # " # 2 1 x2 = − = 1 + x; 1 1 x " # − and so on. We also have the symmetry:

n = n k n k " # " − # The binomial coeﬃcients appear in the expansion

n n (1 + y)2 = yk. = k Xk 0 ! n Likewise, the Gaussian polynomial appear in expansion: k " # 2 n 1 2 n (1 + y)(1 + xy)(1 + x y) (1 + x − y) = 1 + yG (x) + y G (x) + + y G (x) ··· 1 2 ··· n n where G (x) = xk(k 1)/2 k − k " # n Notice that for x = 1, = n . Changing y to yx we get the recursion k k formula stated earlier. " # We now go back to an identity we has porved sometime back:

∞ 2n 1 2 (1 + zx − ) = 1 + zC1(x) + z C2(x) + (1) = ··· Yn 1 7. Lecture 51

where xk2 C (x) = k (1 x2) (1 x2k) − ··· − Now write 64

x2 = X, 1 X = X X2 = X ,..., 1 Xk = X ; − 1 − 2 − k (1 X)(1 X2) (1 Xk) = X, X ... X = X ! − − ··· − 2 k k With this notation, xk2 Ck(x) = Xk! From Jacobi’s triple porduct formula, we have

∞ zl xl2 ∞ 2n 1 1 2n 1 ℓ= (1 + zx )(1 + z x ) = −∞ (2) − − − P n=1 ∞ (1 x2n) Y n=1 − Q By (1), the left side of (2) becomes

∞ r ∞ s ∞ Bn(z, x) z Cr(x) z− Cs(x) = , = = = Xn! Xr 0 Xs 0 Xn 0 where X !is putequalto1. Bn(z, x) is the term correspondingto r+ s = n when the left side◦ is multiplied out in Cauchy fashion. Thus

r s Bn(z, x) = Xn! z − Cr(x)Cs(x) r+s=n X n r2+s2 n 2r x = Xn! z − (r + s = n) Xr!Xn r! r=0 − n X n (n r)2+r2 n 2r = x − z − = r X Xr 0 " # Notice that the powers of z occur with the same parity as n. Now (2) can 65 be re-written as ∞ zl xl2 ∞ Bn(z, x) l= = −∞ X ! P n=0 n ∞ (1 x2n) X n=1 − Q 7. Lecture 52

Both sides are formal power series in x of the appropriate sort. The Bn(z, x) are linear combinations of power series in x with powers of z for coeffieicnts. We can now compare powers of z. We first take only even exponents z2m; we then have infinitely many equations of formal power series. We multiply the 2m m m(m 1) equation arising from z by ( ) x − and add all these equations together; (amd that is the trick, due to Rogers)− we can do this because of linearity. Then

∞ m m(m 1) (2m)2 ( ) x − x ∞ β (x) = − 2l = m 0 , (3) X ! P l=0 2l ∞ (1 x2n) X n=1 − 2l Q 2l 2+ 2 where β (x) = x(2l r) r ( )l r x(l r)(l r 1) 2l r − − − − − = X − Xr 0 " # Writting l r = s, − l 2l 2l2+2s2 s s(s 1) β (x) = x ( ) x − 2l l s − s= l X− " − # l 2 2l 2 = x2l ( )sx3s s l + s − − s= l X− " # (because of the symmetry between l s and l + s). Separating out the term − corresponding to s = 0 and folding together the terms corresponding to s and 66 s, − l 2l2 2l s 2l s(3s 1) 2s β (x) = x + ( ) x − (1 + x ) 2l l − l + s  s=1  " # X " #   l l  2l2  s 2l s(3s 1) s 2l  s(3s+1) = x  ( ) x − + ( )  x − l + s − l + s  s=1 s=0  X " # X " #   l l  2  + 2l + + 2l  + = x2l  ( )s 1 x(s 1)(3s 2) + ( )s  xs(3s 1) (4) − l + s + 1 − l + s  s=0 " # s=0 " #  X X  Then   l l s 2 2l + 1 X + β (x) = x2l ( )s xs(3s 1) 1 − − x4s 2 2l l + s l+s+1 = − − 1 X Xs 0 " # − ! l 2s+1 2 2l + 1 X = x2l ( )s xs(3s 1) − l + s l+s+1 = − 1 X Xs 0 " # − 7. Lecture 53

2l2 l x 2l + 1 + + = ( )s xs(3s 1)(1 x4s 2) (5) 2l+1 l + s + 1 1 X = − − − Xs 0 " # 2m+1 Let us now compute β2l+1(x). For this we compare the coeﬃcients of z , multiply the resulting equations by ( )mxm(m 1) and add up. Then − − ∞ m m(m 1) (2m+1)2 ( ) x − x ∞ β + (x) = − 2l 1 = m 0 , (6) X + ! P l=0 2l 1 ∞ (1 x2n) X n=1 − where Q 67 2l+1 2l + 1 (2l+1 r)2 2 l r (l r)(l r 1) β2l+1(x) = x − + r ( ) − x − − − = r − Xr 0 " # Writting l r = s, this gives − l 2l + 1 (l+1 s)2+(l s)2 s s(s 1) β + (x) = x − − ( ) x − 2l 1 l s − s= l 1 X− − " − # l 2l + 1 2+ + 2+ + 2 = ( )sx3s s l (l 1) l s − s= l 1 X− − " − # l 2+ + 2l + 1 + = x2l 2l 1 ( )s xs(3s 1) − l + s + 1  s=0 " # X  l +  s+1 2l 1 ( s 1)( 3s 2)  + ( ) x − − − − − l + s + 1 s=0  X " #  l  2+ + 2l + 1 + +  = x2l 2l 1 ( )s xs(3s 1)(1 x4s 2) (7) l + s + 1 = − − Xs 0 " # This expression for β2l+1(x) is very neat; it is almost the same as β2l(x) but for trivial factors. Let us go back to β2l+1(x) in its best shape.

2l2+2l+1 2l + 1 β + (x) = x 2l 1 l (" # l 2l + 1 s s(3s+1) 2l + 1 s s(3s 1) + ( ) x + ( ) x − l + s + 1 − l + s − s=1 " # " # ! X  + l + l s+1 2l2+2l+1 2l 1 2l 1 s s(3s 1) 1 X− 2s = x + ( ) x − 1 + −  x l l + s − 1 Xl+s+1 " # s=1 " # − !  X    7. Lecture 54

Since 68 1 Xl s+1 1 Xl+s+1 + Xs Xl+1 (1 Xl+1)(1 + Xs) 1 + − − x2s = − − = − , 1 Xl+s+1 1 Xl+s+1 1 Xl+s+1 − − l+1 − 2l2+2l+1 1 X β + (x) = x − 2l 1 1 X2l+2 − l+1 2l + 2 2l + 2 s s(3s 1) 2s + ( ) x − (1 + x ) l + 1 l + s + 1 − " # s=1 " #   X   ﬀ  This ﬁts with β2l+2. Now we can read o the recursion formulae. The consequences are too very nice facts. The whole thing hinges upon the courage to tackle these sums. We did not do these things ad hoc. Let us compare β2l+1 with β2l

2l+1 2l+1 β + = x (1 X )β ; 2l 1 − 2l l+1 2l 1 1 X β + = x− − − β + ; 2l 1 1 X2l+2 2l 2 − 2l+2 2l+1 1 X so β + = x − β + , 2l 2 1 Xl+1 2l 1 − and β = 1. These things collapse beautifully into something which we could not foresee◦ before. Of course the older proof was shorter. This proof ﬁts very well into our scheme. Lecture 8

Last time we obtained the two fundamental formulae for β2l, β2l+1, from which 69 we deduced the recurrence relations: 2m+1 2(2m+1) β + = x (1 x )β , 2m 1 − 2m 2(2m+2) (1) 2m+1 1 x β + = x − β + 2m 2 1 x2(m+1) 2m 1 − β2m came from B2m by a substitution which was not yet plausible. Let us calculate the ﬁrst few β′ s explicitly. By deﬁnition

B0 = 1 = β0 β = x(1 x2) β = x(1 x2) 1 − 0 − 1 x4 β = x − β = x2(1 x4) 2 1 x2 1 − − β = x3(1 x6) β = x5(1 x4)(1 x6) 3 − 2 − − 1 x8 β = x3 − β = x8(1 x6)(1 x8); 4 1 x4 3 − − − and in general,

2 β = x2m (1 x2m+2)(1 x2m+4) (1 x4m) 2m − − ··· − 2 X ! = Xm 2m (with X = x2); (2) Xm! and similarly,

2m2+2m+1 2m+2 2m+4 4m+2 β + = x (1 x )(1 x ) (1 x ) 2m 1 − − ··· − 2 X + ! = Xm +m x 2m 1 (3) · Xm!

55 8. Lecture 56

This is a very appealing result. We got the β′ s in the attempt of ours to 70 utilise the Jacobi formula. We actually had

∞ l 5l2 l ( ) x − = − ∞ β l 0 = 2m , P X ! ∞ (1 x2m) m=0 2m m=1 − X so that by (2) Q ∞ l l(5l l)/2 ( ) X − 2 = − ∞ Xm l 0 = (4) P X ! ∞ (1 xm) m=0 m m=1 − X Similarly we had Q

∞ ( )lx5l2+3l+1 = − ∞ β + l 0 = 2m 1 , P X + ! ∞ (1 x2m) m=0 2m 1 m=1 − X so that by (3) Q ∞ l l(5l+3)/2 ( ) X + = − ∞ Xm(m 1) l 0 = (5) P X ! ∞ (1 xm) m=0 m m=1 − X Now the right side in theQ Rogers-Ramanujan formula is

∞ 5m 5m 2 5m 3 (1 x )(1 x − )(1 x − ) 1 = m=1 − − − ∞ 5m 1 5m 4 Q ∞ m (1 x − )(1 x − ) (1 x ) m=1 − − m=1 − Q Q which becomes, on replacing x by x2, 71

∞ 10m 10m 4 10m 6 (1 x )(1 x − )(1 x − ) m=1 − − − Q ∞ (1 x2m) m=1 − The numeratoris the same as theQ left side of Jacobi’striple product formula:

∞ 2m 2m 1 1 2m 1 ∞ l l l2 (1 x )(1 zx − )(1 z− x − ) = ( ) z x , − − − − m=1 l= Y X−∞ 8. Lecture 57

with x replaced by x5 and z by x. Hence

∞ 10mm 10m 4 10m 6 ∞ l 5l2+l ∞ l (5l2+l)/2 (1 x )(1 x − )(1 x − ) ( ) X ( ) X l= − − − l= − l= − −∞ = −∞ = −∞ Q P P ∞ (1 x2m) ∞ (1 x2m) ∞ (1 Xm) m=1 − m=1 − m=1 − now Q Q Q

∞ ( )l x5l2+l ∞ ( )kzk xk2 ∞ ( )lxl2+l x(2l)2 l= − k= − ∞ Bn(z, x) l= − −∞ = −∞ = = −∞ , P P X ! P ∞ (1 x2m) ∞ (1 x2m) n=0 n ∞ (1 x2m) m=1 − m=1 − X m=1 −

Q Q+ Q on replacing z2l by ( )lxl(l 1), and this we can do because of linearity. Hence 72 −

∞ l l(5l 1)/2 ( ) X − l= − 1 −∞ = P∞ m ∞ 5m 1 5m 4 (1 X ) (1 x − )(1 x − ) m=1 − m=1 − − Q Q Similarly,

∞ 10m 10m 2 10m 8 (1 x )(1 x − )(1 x − ) 1 = m=1 − − − ∞ 5m 2 5m 3 Q ∞ m (1 X − )(1 X − ) (1 X ) m=1 − − m=1 − Q 2+ Q ∞ ( )lx5l 3l l= − = −∞ . P ∞ (1 Xm) m=1 − Q This time we have to replace z2k+1 by ( )k xk(k 1). Then − − ∞ ( )lXl(5l+3)/2 1 l= − = −∞ ∞ 5m 2 5m 3 P∞ m (1 X − )(1 X − ) (1 X ) m=1 − − m=1 − Q Q These formulae are of extreme beauty. The present proof has at least to do with things that we had already handled. The pleasant surprise is that these things do come out. The other proofs by Watson, Ramanujan and other use 73 8. Lecture 58 completely unplausible combinations from the very start. Our proof is sub- stantilly that by Rogers given in Hardy’s Ramanujan, pp.96-98, though one may not recognize it as such. The proof there contains completely foreign elements, trigonometric functions which are altogether irrelevant here. We now give up formal power series and enter into an entirely diﬀerent chapetr - Analysis. Part II

Analysis

59 Lecture 9

Theta-functions

A power series hereafter shall for us mean something entirely diﬀerent from 74 ∞ n what it did hitherto. x is a complex variable and an x will have a value, its n=0 sum, which is ascertained only only after we introduceP convergence. Then

∞ n f (x) = an x ; = Xn 0 x and the series are coordinatedand we have a functionon thecomplex domain. We take for granted the theory of analytic functions of a complex variable; we shall be using Caushy’s theorem frequently, and in a moment we shall have occasion to use Weierstrass’s double series theorem. Let us go back to the Jacobi identity:

∞ 2n 2n 1 1 2n 1 ∞ k k2 (1 x )(1 + zx − )(1 + z− x − ) = z x − n=1 k= Y X−∞ ∞ k k k2 = 1 + (z + z− )x , (z , 0), = Xk 1 which is a power series in x. Two questions arise. First, what are the domains of convergence of both sides? Second, what does equality between the two sides mean? Formerly, equality meant agreement of the coeﬃcients up to any 75 stage; what it means now we have got to explore. The left side is absolutely convergent - and absolute convergence is enough for us - for x < 1; (for the inﬁnite product (1+a ) is absolutely convergent if a < |;|z is a complex n | n| ∞ Q P 60 9. Lecture 61

variable which we treat as a parameter). For the right side we use the Cauchy- Hadamard criterion for the radius of convergence: 1 ρ = lim √n a | n| = 1 2 lim k z + z k | − | Suppose z > 1; then *****************,p and | | k k k z + z− < 2 z , 2 | k2 | | | and k zk + z k < √2 k z 1 as k | − | | | → →∞ k2 p ∴ lim( zk p+ z k ) 1. | − | ≤ It is indeed = 1, not < 1, becausep ultimately, if k is large enough, z k > 1, and so | | 1 k k k z < z + z− , 2| | | | and we have the reverse inequality. By symmetry in z and 1/z, this holds also for z < 1. The case z = 1 does not present any serious difficulty either. So in all| | cases ρ = 1. Thus| | both sides are convergent for x < 1, and indeed uniformly in any closed circle x 1 δ< 1. | | | |≤ − The next question is, why are the two sides equal in the sense of function 76 theory? This is not trivial. Here equality of values of coefficients up to any definite stage is not sufficient as it was before; the unfinishedcoefficients before multiplication may go up and cannot be controlled. Here, however, we ae in a strong position. We have to prove that

N 2n 2n 1 1 2n 1 ∞ k k k2 (1 x )(1 + zx − )(1 + z− x − ) 1 + (z + z− )x − → n=1 = Y Xk 1 with increasing N, when x < 1, and indeed uniformly so in x 1 δ < 1. On the left side we have a| sequence| of polynomials: | | ≤ −

N = 2n + 2n 1 + 1 2n 1 = ∞ (N) m fN (x) (1 x )(1 zx − )(1 z− x − ) am x , say. = − = Yn 1 Xm 0 (of course the coefficients are all zero beyond a certain finite stage). Now we know that the left side is a partial product of a convergent infinite product; in fact fN (x) tends uniformly to a series, f (x), say. Now what do we know about 9. Lecture 62

a sequence of analytic functions on the same domain converging uniformly to a limit function? The question is answered by Weierstrass’s double series theorem. We can assert that f (x) is analytic in the same domain at least, and ∞ m further if f (x) = am x , then m=0 P = (N) am lim am . N →∞ The coefficients of the limit function have got something to do with the 77 original coefficients. Now 1 f (x) a(N) = N dx m 2πi xm+1 x =Z1 δ | | − (N) Let N ; this is permissible by uniform convergence and the am , s in fact converge→ ∞to 1 f (x) a = dx. m 2πi xm+1 x =Z1 δ | | − (Weirestrass’ own proof of this theorem was what we have given here, in some disguise; he takes the values at the roots of unity and takes a sort of mean value). k k k2 Now what are the coefficients in 1 + (z + z− )x ? Observe that the con- (N) (N) vergence of am to am is a peculiar and simple one. am indeed converges to a (N) P known am; as a matter of fact am = am for N sufficiently large. They reach a limit and stay put. And this is exactly the meaning of our formal identity. So the identity has been proved in the function-theoretic sense:

∞ 2n 2n 1 1 2n 1 ∞ k k k2 ∞ k k2 (1 x )(1 + zx − )(1 + z− x − ) = 1 + (z + z− )x = z x . − n=1 k=1 k= Y X X−∞ These things were done in full extension by Jacobi. Let us employ the ususl 78 symbols; in place of x write q, q < 1, and put z = e2πiv. Notice that the right side is a Laurent expansion in z| in| 0 < z < (v is unrestricted because we hace used the exponential). We write in the| | traditional∞ notation

∞ 2n 2n 1 2πiv 2n 1 2πiv (1 q )(1 + q − e )(1 + q − e− ) = − Yn 1 ∞ 2 = qn e2πinv n= X−∞ 9. Lecture 63

= v3(v, q) v3 (and in fact all the theta functions) are entire functions of v. We have taken q < 1; it is customary to write q = eπiτ, so that q < 1 implies | | | | eπiτ = eRπiτ, Rπiτ< 0 | | i.e., Riτ< 0 or I mτ> 0

τ is a point in the upper half-plane. τ and q are equivalent parameters. We also write V3(V , q) = V3(V /τ) (An excellent accout of the V -functionscan be found in Tannery and Molk: Fonctiones Elliptiques, in 4 volumes; the second volume contains a very well orginized collection of formulas). One remark is immediate from the deﬁnition of V3, viz.

V3(V + 1, q) = V3(V , q)

On the other hand, 79

∞ 2n 2n 1 2πiV 2πiτ 2n 1 2πiV 2πiτ V3(V + τ, q) = (1 q )(1 q − e e ) (1 + q − e− e− ) = − − × Yn 1 ∞ 2 = qn e2πinV e2πinV , n= X−∞ and since q = eπiτ,

∞ 2n 2n+1 2πiV 2n 3 2πiV ∞ n2+2n 2πinV (1 q )(1 + q e )(1 + q − e− ) = q e − n=1 n= Y X−∞ or

V 1 + q 1e 2πi ∞ − − (1 q2n)(1 + q2n 1e2πiV )(1 + q2n 1e 2πiV ) + 2πiV − − − 1 qe = − Yn 1 1 2πiV ∞ (n+1)2 2πi(n+1)V = q− e− q e n= −∞ 1 2πiV X = q− e− V3(V , q) 2πiV 1 = (qe )− V3(V , q) 9. Lecture 64

So we have the neat result:

1 2πiV V3(V + τ, q) = q− e− V3(V , q)

1 is a period of V3 and τ resembles a period. It is quite clear that we cannot 80 expect 2 peroids in the full sense, because it is imposible for an entire function to have two periods. Indeed if ω1 and ω2 are two periods of f , then f (V + ω1) = f (V ), f (V + ω2) = f (V ), and f (V + ω1 + ω2) = f (V ) and the whole module generated by ω1 and ω2 form periods. Consider the fundamental region which is the parallelogram with vertices at 0, ω1, ω2, ω1 + ω2. If the function is entire it has no poles in the parallelogram and is bounded there (because the parallelogram is bounded and closed), and therefopre in the whole plane. Hence by Liouville’s theorem the function reduces to a constant. While dealing with trigonometric functions one is not always satisﬁed with the cosine function alone. It is noce to have another function: cos(x π/2) = sin x. A shift by a half-period makes it concenient for us. Let us− consider V V + 1 V V + V V + 1 + τ analogously 3( 2 , q), 3( τ/2, q), and 3( 2 2 , q). Though τ is not strictly a period we can still speak of the funcdmental region, because V V + 1 on shifting by τ we change only by a trivial factor. Replace by 2 and everything is ﬁne as 1 is a period.

1 ∞ 2n 2n 1 2πiV 2n 1 2πiV V3(V + , q) = (1 q )(1 q − e )(1 q − e− ) 2 = − − − Yn 1 ∞ 2 = ( )nqn e2πinV n= − X−∞ which is denoted V4(V , q) Again 81

τ ∞ 2n 2n 1 2πiV πiτ 2n 1 2πiV πiτ V3(V + , q) = (1 q )(1 + q − e e )(1 + q − e− e− ) 2 = − Yn 1 ∞ 2 = qn e2πinV eπinτ n= X−∞ 2πiV ∞ 2n 2n 2πiV 2n 2πiV i.e., (1 + e− ) (1 q )(1 + q e )(1 + q e− ) = − Yn 1 ∞ 2 = qn +ne2πinV n= X−∞ 9. Lecture 65

1/4 πiV ∞ (n+1/2)2 (2n+1)πiV = q− e− q e n= −∞ 1/4 πiV X = q− e− V2(V , q)

∞ (n+1/2)2 (2n+1)πiV 1/4 where V2(V , q) = q e , by deﬁnition. (Here q− does not n= −∞ πiτ/4 contain an unknown4thP root of unity as factor, but is an abbreviation for e− , so that it is well deﬁned). So

1/4 ∞ 2n 2n 2πτV 2n 2πiV V2(V , q) = 2q cos πV (1 q )(1 + q e )(1 + q e− ) = − Yn 1 Finally 82 + 1 τ 1/4 πi(V + 1 ) 1 V (V + , q) = q− e− 2 V V + , q 3 2 2 2 ! 1/4 1 πiV 1 = q− e− V V + , q i 2 2 ! 2 1/4 πiV ∞ n ( 2n+1 ) (2n+1)πiV = q e− ( ) q 2 e n= − X−∞ 2 1 πiV = cos π V + e− i 2 ! ∞ 2n 2n 2πiV 2n 2πiV (1 q ) 1 q e 1 q e− − − × − n=1 Y Now deﬁne 1 V (V , q) = V V + , q , 1 2 2 ! or

1/4 ∞ 2n 2n 2πiV 2n 2πiV V1(V , q) = 2q sin πV (1 q )(1 + q e )(1 q e− ) = − − Yn 1 2 1/4 ∞ n ( 2n+1 ) (2n+1)πiV = iq− ( ) q 2 e m= − X−∞ Collecting together we have the four V -functions: 83

2 1/4 ∞ n ( 2n+1 ) (2n+1)πiV V1(V , q) = iq− ( ) q 2 e m= − X−∞ 9. Lecture 66

2 ∞ n 2n+1 = ( ) q( 2 ) sin(2n + 1)πV = − Xn 0 ∞ ( 2n+1 )2 V2(V , q) = 2 q 2 cos(2n + 1)πV = Xn 0 ∞ n2 V3(V , q) = 1 + 2 q cos2nπV = Xn 1 ∞ n n2 V4(V , q) = 1 + 2 ( ) q cos2πnV = − Xn 1

Observe that the sine function occurs only in V1. Also if q, V are rel these reduce to trigonometric expansions. Lecture 10

Let us recapitulate the formulae we has last time. 84

2 1 ∞ n 2n+1 (2n+1)πiV V (V , q) = ( ) q( 2 ) e 1 i n= − X−∞ 2 ∞ n 2n+1 = 2 ( ) q( 2 ) sin(2n + 1)πV n= − X∞ 1/4 ∞ 2m 2m 2πiV 2m 2πiV = 2q sin πV 1 q 1 q e 1 q e− (1) − − − m=1 Y 2 ∞ ( 2n+1 ) (2n+1)πiV V2(V , q) = q 2 e n= X−∞ ∞ 2n+1 2 = 2 q( 2 ) cos(2n + 1)πV = Xn 0 1/4 ∞ 2m 2m 2πiV 2m 2πiV = 2q cos πV 1 q 1 + q e 1 + q e− (2) − m=1 Y ∞ n2 2πiV V3(V , q) = q e n= X−∞ ∞ 2 = 1 + 2 qn cos2nπV = Xn 1 ∞ 2m 2m 1 2πiV 2m 1 2πiV = 1 q 1 + q − e 1 + q − e− (3) − m=1 Y

67 10. Lecture 68

∞ n n2 2nπiV V4(V , q) = ( ) q e n= − X−∞ ∞ 2 = 1 + 2 ( )nqn cos2nπV = − Xn 1 ∞ 2m 2m 1 2πiV 2m 1 2πiV = (1 q )(1 q − e )(1 q − e− ) (4) = − − − Ym 1

We started with V3 and shifted the argument V by ‘periods’, and we had, 85 writing q = eπiτ, V3(V + 1, q) = V3(V , q) 1 2πiV (5) V3(V + τ, q) = q− e− V3(V , q). Then we took ‘half-periods’ and then something new happened, and we gave names to the new functions:

1 V V + , q = V (V , q) 3 2 4 ! τ 1/4 2πiV V V + , q = q− e− V (V , q) (6) 3 2 2 + 1 τ 1/4 πiV V V + , q = iq− e− V (V , q) 3 2 1 ! Let us study how these functions alter when the argument V is changed by 1, τ,1/2, τ/2, (1+τ)/2. V V +1 is trivial; V V +1/2 is also easy to see by inspection. Let us take V→+ τ. (We suppress the→ argument q for convenience of writing).

1 1 + τ V (V ) = q1/4e2πiV V V + 1 i 3 2 ! 1 1 + τ ∴ V (V + τ) = q1/4eπi(V +τ)V V + τ + 1 i 3 2 ! + 1 1/4 πiV 1 2πi(V +1+τ/2) 1 τ = q e qq− e− V V + i 3 2 ! 2πiV πi(1+τ) = e− e− V1(V , q) = AV (V , q), − 1 1 2πiV where A = q− e− ; the other conspicuous factor which occurs in similar 86 1/4 2πiV contexts is denoted B = q− e− . 10. Lecture 69

The other transformations can be worked out in a similar way by ﬁrst going over to V3. We collect the results below in tabular form.

V + V + V + 1 V + 3 V + 1+τ 1 τ 2 2 2

V V AV V iBV BV 1 − 1 − 1 2 4 3 V V AV V BV iBV 2 − 2 2 − 1 3 − 4

V3 V3 AV3 V4 BV2 BV1

V V AV V iBV BV 4 4 − 4 3 1 2 It may be noticed that each column in the table contains all the four functions; so does each now. The systematique of the notation for the V -functions is rather questionable. Whittaker and Watson write V instead of πV , which has the unpleasant consequence that the ‘periods’ are then π and πτ. Our notation is the same as in 87 Tannery and Molk. An attempt was made by Kronecker to systematise a little the unsystematic notation. Charles Hermite introduced the following notation:

+ 2 ∞ νn 2r µ (2n+µ)πiV Vµν(V , q) = ( ) q 2 e n= − X−∞ + 2 ∞ νn 2n µ πiτ (2n+µ)πiV = ( ) e 2 e n= − X−∞ where µ, ν = 0, 1 . In this notation, ∗ ∗ ∗ ∗∗

V00(V , q) = V3(V , q)

V01(V , q) = V4(V , q)

V10(V , q) = V2(V , q)

V11(V , q) = iV1(V , q).

This, however, has not found any followers. While writing down derivatives, we always retain the convention that a prime refers to diﬀerentiation with respect to V :

∂ V (V , q) = V (V , q)(α = 1, 2, 3, 4) α′ ∂ν α 10. Lecture 70

Taking partial derivatives, we have

+ 2n+µ 2 ∂ ∞ νn 2n µ πiτ (2n+µ)πiV V (V /τ) = ( ) πi e 2 e , ∂τ µν 2 n= − X−∞ ! 2 + 2 ∂ ∞ νn 2n µ πiτ 2 2 2 (2n+µ)πiV and V (V /τ) = ( ) e 2 π i (2n + µ) e , ∂V 2 µν n= − X−∞ Comparing these we see that they agree to some extent; in fact, 88

∂ ∂2 4πi V (V /τ) = V (V /τ) (7) ∂τ µν ∂ν2 µν This is a partial differential equation of the second order, a parabolic equation with constant coefficients. It is fundamental to write iτ = t; (7) then becomes the differential equation for heat conduction. V -functions− are thus very useful tools in applied mathematics; they were used by Poisson and Fourier in this connection. Again,

2 + 2n+µ 2 ∂ ∞ ν 2n µ 1 (2n+µ)πiV V (V , q) = ( ) q 2 − e , ∂q µν 2 n= − −∞ ! X 2 2 ∂ ∂ 4π q Vµν(V , q) = Vµν(V , q), (8) − ∂q ∂ν2 which is another form of (7). Here the uniformity of notation was helpful; it was not necessary to discuss the diﬀerent functions separately. We now pass on to another important topic. The zeros of the theta - functions. The V -functions are more or less periodic. The exponential factor that is picked up on passing from one parallelogram to another is non-zero and can accumulate. It is evident from the deﬁnition that

V , (0, q) = 0.

On the other hand V2, V3, V4 , 0. (when the argument V is 0 we write 89 hereafter simply V ). This is so because the inﬁnite products are absolutely 1 1 convergent. (Let us recall that a product like 1 2 3 is not properly convergent in the product sense). Again from the deﬁnitions,· · ···

1 V = 0 2 2 ! 10. Lecture 71

1 + τ V = 0 3 2 ! τ V = 0 4 2 So far we have one zero per parallelogram for each of the functions; and there can be no other in a parallelogram, as can be seen from the inﬁnite product expansions. The zeros of V1(V , q) are m1 + m2τ(m1, m2 integers), for 1 e2πimτe2πiV = 0 implies mτ + V = m or V = m mτ. The zeros − 1 1 1 − of V1(V , q), V2(V , q), V3(V , q), V4(V , q) in the fundamental parallelogram are 1 1+τ τ nicely arranged in order at the points 0, 2 , 2 , 2 respectively.

0 1

All the zeros are therefore given by the formulae: 90

V1(m1 + m2τ) = 0 1 V m + m τ + = 0 2 1 2 2 ! 1 + τ V m + m τ + = 0 3 1 2 2 ! τ V m + m τ + = 0 4 1 2 2 It is of interest to study Vα(0, q) (usually written Vα).

V1(0) = 0

2 ∞ (2n+ 1 ) V2(0) = q 2 n= X−∞ ∞ = 2q1/4 (1 q2m)(1 + q2m)2 = − Ym 1 10. Lecture 72

= V2

∞ n2 V3(0) = q n= X−∞ ∞ 2m 2m 1 = (1 q )(1 + q − ) = − Ym 1 ∞ n n2 V4(0) = ( ) q n= − X−∞ ∞ 2m 2m 1 2 = (1 q )(1 q − ) = − − Ym 1 We cannot anything of interest in V1. Let us look at the others. 91

∞ 2n+1 2 V = V = n + ( 2 ) 1′(0, q) 1′ 2π ( ) (2n 1)q = − Xn 0 ∞ ∞ ′ = 2q1/4 π cos πV ( ) + sin πV ( ) ··· ···  m=1 m=1  V =0  Y Y    ∞    = 2πq1/4 (1 q2m)3    = − Ym 1 Immediately we see that this yields the interesting identity of Jacobi.

∞ ∞ 2 (1 q2m)3 = ( )n(2n + 1)qn +n, = − = − Ym 1 Xn 0 or, replacing q2 by x,

∞ ∞ (1 xn)3 = ( )n(2n + 1)xn(n+1)/2 = − = − Yn 1 Xn 0 We had proved this earlier by the method of formal power series. Here we can diﬀerentiate with good conscience. Now 2 ∞ 2m 2m 1 2m 1 πV V V = V ′ (1 + q )(1 + q − )(1 q − ) 2 3 4 1 − m=1  Y   2   ∞ 2m 4m 2  = V ′ (1 + q )(1 q − ) , 1 − m=1  Y     10. Lecture 73

which becomes, on replacing q2 by x, 92

2 ∞ n 2n 1 V ′ (1 + x ) 1 x − 1 − m=1  Y    ∞ n  2n 1  However, (1 + x )(1 x − ) = 1. We therefore have the very useful and m=1 − pleasant formulaQ V = V V V 1′ π 2 3 4 Lecture 11

We found that Vα(V , q) changes at most its sign when V is replaced by V + 1, 93 while it picks up a trivial factor A when V is replaced by V + τ. If we form quotients, A will cancel out and we may thereforeexpect to get doubly-periodic functions. Let us form some useful quotients:

V2(V , q) f2(V ) = V1(V , q) V3(V , q) f3(V ) = V1(V , q) V4(V , q) f4(V ) = V1(V , q)

For simplicity of location of poles it is convenient to take V1 in the denominator since it has a zero at the origin. From the table of the V -functions we ﬁnd that these functions are not quite doubly periodic: f (V + 1) = f (V ) f (V + 1) = f (V ) 2 2 3 − 3 f (V + τ) = f (V ) f (V + τ) = f (V ) 2 − 2 3 − 3 f (V + 1) = f (V ) 4 − 4 f4(V + τ) = f4(V )

So the functions are not doubly periodic; they do not return to themselves. 94 And we cannot expect that either. For suppose any of the functions f were actually doubly periodic. We know that each has a pole of the ﬁrst order per 1+τ 1 τ parallelogram. Integrating round the parallelogram with vertices at 2 , −2 (so that the origin which is the pole is enclosed), we have ± ±

f (V )dV = 0 Z 74 11. Lecture 75

0 1

i.e., the sum of the residues at the poles =0. This means that either the pole is a double is a double pole with zero residue, or there are two simple poles with residues equal in magnitude but opposite in sign. However neither of these is the case. So there is no necessity for any further experimentation. Let us therefore consider the squares 2 V 2 V 2 V f2 ( ), f3 ( ), f4 ( ) these are indeed doubly periodic functions. And they are even functions. So the expansion in the neighbourhood of the pole will not contain the term of power 1. Hence the pole must be a double pole with residue zero. So they are closely− related to the Weierstrassian function P(V ), and must indeed be of the form CP(V ) + C1. So we have constructed doubly periodic functions. They are essentially 95 P(V ). ω1 and ω2 of P(V ) are our 1 and τ. In order to get a better insight we need the exact values of the functions. Let us consider their pole terms. Expanding in the neighbourhood of the origin, V V V V + α′′ V 2 + α( , q) = α 2! V V ··· V1(V , q) 1′ V + 1′′′ V 3 + 1! 3! ··· V 1 α′′ 2 V 1 + V ν + = α 2 α ··· V νV ′ 1′′′ 2 1  1 + V ν +   6 1′   ···    V 2 Vα  1 V ′′  ′′′ = 1 + α ν2 +  1 1 ν2 + + ( )3 νV 2 V ··· − 6V ··· ··· −··· 1′ α !  1′ !    V V V ′′′   = α 1 + ν2 ′′ 1  +  νV 2V − 6V ··· 1′ α 1′ ! ! V (V , q) 2 ∴ f 2 = α α V (V , q) 1 ! 11. Lecture 76

2 V 1 V V ′′′ = α 1 + V 2 α′′ 1 + V 2 V 2 V − 3V ··· ′ 1 1 1′ ! !

Let us now specialise α. We have a special interest in V3 because it is such a V ′2 2 1 V = ∞ n 1 2 V = + nice function: 3 q . We have 2 fα ( ) non-negative powers n= V V2 −∞ α of V . P If we take two such and take the diﬀerence, the diﬀerence will no longer 96 have a pole. Taking α = 2, 4, for instance,

2 2 2 2 V ′ V (V , q) V ′ V (V , q) V ′′ V 1 2 1 4 = 2 4 +positive powers of V (*) V 2 V (V , q) − V 2 V (V , q) V − V 2 1 ! 4 1 ! 2 4′′ The left side is a doubly periodic function without a pole and so a constant V ′′ V ′′ C; the right side is therefore just 2 4 . The vanishing of the other terms V2 − V4 on the other terms on the right side, of course, implies lots of identities. So we have already computed C in one way:

V ′′ V ′′ C = 2 4 V2 − V4 1 τ To evaluate C in other ways we may take in (*) V = , V = or V = 2 2 (1 + τ)/2. From the table, + 1 1 τ 1/4 V , q = V V , q = q− V 1 2 2 1 2 3 ! ! + 1 1 τ 1/4 V , q = V = 0 V , q = iq− V 2 2 − 1 2 2 − 4 ! ! + 1 1 τ 1/4 V , q = V V , q = q− V 4 2 3 4 2 2 ! ! So again from the left side of (*), 97

V ′2 V 2 V 2 C = 1 0 1 3 V 2 V 2 V 2 2 × − 4 2 π2V 2V 4 = 1 3 = π2V 4 2V 2V 2V 2 3 −π 2 3 4 − 11. Lecture 77

Also

V ′2 V 2 V ′2 V 2 C = 1 4 1 2 V 2 −V 2 − V 2 V 2 2  3  4 3 π2V ′2V 4 π2V ′2V 4 =  1 4 1 2 2V 2V 2V 2 2V 2V 2V 2 −π 2 3 4 − π 2 3 4 = π2V 4 π2V 4 − 4 − 2 From these we get an identity which is particularly striking:

V 4 = V 4 + V 4 3 2 4 (1) We have also V V 2V 4 = 4′′ 2′′ π 3 (2) V4 − V2 Now let us look at (1) and do a little computing. Explicitly (1) states:

4 4 4 ∞ 2 ∞ ∞ 2 qn = q1/4 qn(n+1) + ( )nqn (3) − n=   n=  n=  X−∞ X−∞ X−∞       This is an identity of some interest.    Let us look for qN on both sides. The left side gives N in the form N = 98 2 + 2 + 2 + 2 n n2 n3 n4, that is, as the sum of four squares. So dies the second term on the right. If N is even, it is trivial that both sides are in agreement because the ﬁrst term on the right gives only odd powers of q, and the coeﬃcient of qn in the second term on the right is

+ + + ( )n1 n2 n3 n4 − n2n2+n2+n2=N 1 2 X3 4

Since N is even either all ni’s are odd, or two of them odd,or none. It is not transperent. What happens when N is odd. Take the more interesting formula (2): V V V 4 = 4′′ 2′′ π 3 V4 − V2 By the diﬀerential equation,

∂2 V ′′ = V (V , q) α ∂V 2 α " #V =0 11. Lecture 78

2 ∂ = 4π q Vα(V , q) − ∂q " #V =0 1 ∂V 1 ∂V ∴ V 4 = 4q 2 4 3 V ∂q − V ∂q 2 4 ! ∂ V = 4q log 2 ∂q V4

Now 99

∞ 2n + 2n 2 V (1 q )(1 q ) 2 = 1/4 n=1 − V 2q 4 ∞Q 2n 2n 1 (1 q )(1 q − ) n=1 − − Q ∞ (1 q2n)2(1 + q2n)2 = − = 2q1/4 n 1 ∞Q 2n 2 2n 1 2 (1 q ) (1 q − ) n=1 − − Q ∞ (1 q4n)2 = − = 2q1/4 n 1 Q ∞ (1 qn)2 n=1 − Q1 = 2q /4 (1 qn)2 4 ∤ n − Q Taking the logarithmic derivative,

1 nqn 1 V 4 = 4q 2 − − 3 4q − 1 qn  4 ∤ n −  X   nqn  = 1 +8   1 qn  ∤ 4Xn − ∞ = 1 + 8 n qnk ∤ = 4Xn Xk 1 = 1 + 8 qm n 4 ∤ m n m Xm=1 X|

∞ m = 1 + 8 σ∗(m)q = Xm 1 11. Lecture 79 with the previous that σ∗(m) = , d , that is the divisor sum with those 100 4 ∤ α d/m divisors omitted which are divisibleX by 4. This is an interesting identity:

4 ∞ n2 ∞ m q = 1 + 8 σ∗(m)q (4) n=  m=1 X−∞ X   m   n2+n2+n2+n2 On the left q can be obtained only as q 1 2 3 4 , so that the coeﬃcient of qm on the right is the number of ways in which this representation for m is possible; m is as often the sum of four squares as 8σ∗(m). Clearly σ∗(m) , 0, since among the admissible divisors, 1 is always present. So σ∗(m) 1, or every m does admit at least one such representation. We have thus≥ proved Lagrange’s theorem: Every integer is the sum of at most four squares. If m is odd, σ∗(m) = σ(m); if m is even,

σ∗(m) = d + 2 d d m, d odd d m, d odd | X | X = 3 d d m, d odd | X If we denote by r4(m) the number of representations of m as the sum of four squares, then r4(m) = 8 times the sum of odd divisors of m, m odd; 24 times the sum of odd divisors of m, m even.

We have not partitions this time, but representation as the sum of squares. We agree to consider as distinct these representations in which the order of the components has been changed. In partitions we abstracted from the order of 101 the summands; here we pay attention to order, and also to the sign (i.e., one 2 + 2 + 2 + 2 ff representation n1 n2 n3 n4 is actually counted, order apart, as 16 di erent 2 2 2 2 representations ( n1) +( n2) +( n3) +( n4) , if n1, n2, n3, n4 are all different from 0). ± ± ± ± As an example, take m = 10. The different representations as the sum of four squares are

( 1)2 + ( 1)2 + ( 2)2 + ( 2)2, ± ± ± ± ( 1)2 + ( 3)2 + (0)2 + (0)2, ± ± along with their rearrangements, six in each. Thus altogether

r (10) = 6 16 + 6 8 = 144 4 × × 11. Lecture 80

8σ∗(10) = 3(1 + 2 + 5 + 10) = 8 18 = 144 × Lagrange’s theorem was ﬁrst enunciated by Fermat in the seventeenth cen- tury. Many mathematicians tried to solve it without success; eventually Jacobi found out the identity r4(m) = 8σ∗(m) Before that, the fact that every integer is the sum of four squares was conjectured by Fermat, Euler did not succeed in proving it. It was proved by La- grange, and later Euler gave a mere elementary proof. Euler proved that if two numbers are each the sum of four squares, then so is their product, by means of the identity:

2 + 2 + 2 + 2 2 + 2 + 2 + 2 (x1 x2 x3 x4)(y1 y2 y3 y4) = (x y + x y + x y + x y )2 + (x y x y + x y x y )2+ 1 1 2 2 3 3 4 4 1 2 − 2 1 3 4 − 4 3 + (x y x y + x y x y )2 + (x y x y + x y x y )2. 1 3 − 3 1 4 2 − 2 4 1 4 − 4 1 2 3 − 3 2 We do not proceed to discuss in detail the representability of a number as 102 the sum of two sequences. 2 If we return not to fα but to fα we are not helpless to deal with them. f4 is not doubly periodic in the fundamental parallelogram, but is doubly periodic in a parallelogram of twice this size with vertices at 0, 2, 2 + τ, τ. Ithas gotapole at the vertex 0 and another at the vertex 1, with residues adding up to zero.

0 1 2

We may write down another identity:

V τ V +1 τ V V V V ′ V ′ 1′ 4( /τ) = 1 1 2 2 1 2 2 V V + V4 · V1(V /τ) 2  V .τ − V 1 .τ   1 2 2 1 2 2     . .  This may be deduced by checkingthat the poles on both sides are the same, Further they are odd functions and so the constant term in the diﬀerence must 1 vanish. Put V = on both sides. 2 11. Lecture 81

Then we get 103 1 τ 3 τ V ′ V ′ V 2 = 1 1 4 2 1 4 2 π 3 2  V 1 .τ − V 3 .τ   1 4 2 1 4 2    By straightforward calculation, taking . logarithmic . deriv atives, we obtain,  

V 2 = ∞ m (1) (3) 3 4 q (σ (m) σ (m)), = ◦ − ◦ Xm 1 where the notation employed is:

k σk(m) = d , d m X| σ (m) = d◦ = number of divisors of m; ◦ d m X| ( j) σ (m) = d◦ ◦ d m, d j (mod 4) | X≡ comparing coefficients of qm, and observing that on the left m occurs only in 2 + 2 the form n1 n2, we get the beautiful theorem: m can be represented as the sum of two squares as often as 4(σ(1)(m) σ(3)(m)). ◦ − ◦ Notice that σ(1)(m) σ(3)(m) is always non negatives; hence σ(1)(m) σ(3)(m) (i.e., the number◦ − of◦ divisors of the form 4r + 1 is never less◦ than the≥ number◦ of divisors of the form 4r + 1), which is by no means a trivial fact. (1) (3) In some cases we can actually find out what the difference σ (m) σ (m) 104 will be. Suppose that m is a prime p. Then the only divisors◦ are− 1 and◦ p. The divisor 1 goes into σ(1); and p goes into σ(3) if p 3 (mod 4). So the difference is zero. However,◦ if p 1 (mod 4).◦ p goes≡ into σ(1). Hence the number of representations of a prime≡ p 1 (mod 4) as thesum of◦ two squares is 4 2 = 8. That the number of representations≡ of a prime p 1 (mod 4)as the sum× of two squares is 8 is a famous theorem of Fermat, prove≡ d for the first time by Euler. It is usually proved by using the Gaussian complex numbers. So far we have been looking upon V as the variable in the V - functions; now we proceed to consider q as the variable and go to deeper things like the Jacobi transformation. Lecture 12

We now come to a rather important topic, the transformation of V -functions. 105 So far we have been looking upon Vα(V /τ) as a function of V only; hereafter we shall be interfering with the ‘period’ τ also. We want to study how Vα(V /τ) changes when V is replaced by V + 1/τ. For this it is enough if we replace V by V τ = ω and see how the function behaves when ω is changed to ω+1. This would amount to turning the whole plane around in the positive sense about the origin through arg τ. We take V1, because it is easier to handle, since the zeros become the periods too. Consider

f (V ) = V1(V τ/τ)

Then

f (V + 1) = V1((V + 1)τ/τ)

= V1(V τ + τ/τ) πiτ 2πiV τ = e− e− V (V τ/τ) − 1 πiτ 2πiV τ = e− e− f (V )

V 1 1 τ Similarly consider f ( 1/τ) (We choose to take τ rather than τ since we want the imaginary part of− the parameter to be positive:− 1 τ¯ 1 Im = Im < 0andso Im > 0) τ ττ¯ − τ 1 1 f (ν ) = V ((ν )τ/τ) − τ 1 − τ = V (V τ 1/τ) 1 − = V (V τ/τ) − 1

82 12. Lecture 83

= f (V ) − So f is a sort of V -function which picks up simple factors for the ‘periods’ 1 106 and 1 . f (V ) hasclearly zerosat0 and τ = 1 , or generallyat V = m +m τ ; − τ ′ − τ 1 2 ′ m1, m2 integers, which is a point-lattice similar to the old one turned around. Similarly let us deﬁne

1 g(V ) = V (V /τ′) = V V / 1 1 − τ ! g(V + 1) = V1(V + 1/τ′)

= V (V /τ′) − 1 = g(V ) − 1 g V = g(V + τ′) − τ ! = V1(V + τ′/τ′) πiτ 2πiV = e− ′ e− V (V /τ′) − 1 πi/τ 2πiV = e e− g(V ) − Let us form the quotient: 107

f (V ) Φ(V ) = g(V ) f (V + 1) φ(V + 1) = g(V + 1) πiτ 2πiV τ = e− e− φ(V ) − V + τ Φ V 1 = f ( ′) − τ g(V + τ ) ! ′ V = f ( ) πi/τ 2πiV e e− g(V ) πi/τ 2πiV = e− e Φ(V )

Φ takes on simple factors in both cases of this peculiar sort that we can eliminate them both at one stroke. We write

πiτ(2V +1) e− Φ(V ) =Φ(V + 1), πi(2V 1/τ) e − Φ(V ) =Φ(V 1/τ) − 12. Lecture 84

Let us try the following trick. Let us supplement Φ(V ) by an outside function h(V ) so that the combined function Φ(V ) is totally doubly periodic. Write

Ψ(V ) =Φ(V )eh(V )

We want to choose h(V ) in such a manner that 108

Ψ(V + 1) = Ψ(V + τ′) = Ψ(V )

This implies two equations:

πiτ(2V +1) h(V +1) h(V ) e− e − = 1 πi(2V 1τ) h(V 1/τ) h(V ) e − e − − = 1;

h(V + 1) h(V ) = πiτ(2V + 1) + 2πim − h(V + τ′ h(V ) = πi(2V + τ′) + 2πim′. − − We can solve both at one stroke. Since on the right side we have a linear function of V in both cases, a quadratic polynomial will do what we want.

(V + δ)2 V 2 = 2V δ + δ2 = δ(2V + δ), − and taking h(V ) = πiτV 2,

h(V + 1) h(V ) = πiτ(2V + 1) − h(V + τ′) h(V ) = πiττ′(V + τ′) = πi(2V + τ′), − − so that both the equations are satisﬁed. Putting it in, we have

2 V (V τ/τ) Ψ(V ) = eπiτV 1 V V / 1 1 − τ This has the property that 109

ψ(V + 1) = ψ(V + τ) = ψ(V )

So we have double periodicity. This function is also an entire function because the numerator and denominator have the same simple zeros. So this is a pole-free function and hence a constant C, C a constant with respect to the variable V , but may be a function of the parameter τ, C = C(τ). We thus have 12. Lecture 85

I. 2 1 eπiτV V (V τ/τ) = C(τ)V V / 1 1 − τ ! What we need now are the corresponding formulas for the other functions. V V + 1 Replacing by 2 ,

πiτ(V + 1 )2 1 1 1 e 2 V V + τ/τ = C(τ)V V + , 1 2 1 2 − τ ! ! . ! πiτ(V 2+V +1/4) πiτ/4 πiV τ 1 or e ie− e− V (V τ/τ) = C(τ)V V / 4 2 − τ ! We notices here that two diﬀerent V -functions are related. This gives II. 2 1 ieπiτV V (V τ/τ) = C(τ)V V / . 4 2 − τ ! Replacing in I V by V + τ′/2 = V 1/(2τ), we get III. − 2 1 ieπiτV V (V τ/τ) = C(τ)V V / 2 4 − τ ! V + 1 V Finally putting 2 for In, III, IV. 2 1 ieπiτV V (V τ/τ) = C(τ)V V / 3 3 − τ ! The way the functions change over in I-IV is quite plausible. For consider 110 the location of the zeros. When we take the parallelogram and turn it around what was originally a zero for V4 becomes 3 one for V2 and vice 4 versa; and what used to be in the middle, the zero of 1 2 V3, Remains in the middle. So the formulae are plausible in structure. The most important thing now is, what is C(τ)? To evaluate C(τ) let us diﬀerentiate I and put V = 0. We have 12. Lecture 86

V. 1 τV 1(0/τ) = C(τ)V 1 0/ 1 1 − τ ! From II, III, and IV, putting V = 0,

1 iV (0/τ) = C(τ)V 0/ 4 2 − τ ! 1 iV (0/τ) = C(τ)V 0/ 2 4 − τ ! 1 iV (0/τ) = C(τ)V 0/ 3 3 − τ ! V = V V V Multiplying these together and recalling that π 1′ 2 3 4, we obtain VI. 1 3 1 iV (0/τ) = (C(τ)) V ′ 0/ . − 1 1 − τ ! Dividing by VI, by V, 111 1 = C2(τ), iτ 1 or C(τ) = ± riτ C(τ) In II, III, IV, it is i that appears; so let us write this is

C(τ) = 1 1 = i i ± i riτ ± rτ C(τ) Now k(i/τ) > 0 i is completely determined, analytically, in particular by IV: · V πiτn2 C(τ) = 3(0/τ) = e 1 2 i V (0/ ) eπiτ′n 3 − τ P Both the numerator and denominator are analyticP functions if Im τ> 0. So C(τ) i is analytic and therefore continuous. i/τ must lie in the right half-plane, i and thus τ in either of the sectors with central angle π/2, but because of continuityq it cannot lie on the border lines. So it is in the interior of entirely one sector. To decide which one it is enough if we make one choice. 12. Lecture 87

Take τ = it, t > 0; then 112

πtn2 C(it) = e− i e (π/t)n2 P − P C(τ) Both numerator and denominator are positive. So i lies in the right half. So arg i < π and i denotes the principal branch. The last equality gives: | τ | 4 τ q q ∞ πin2/τ τ ∞ πin2τ e− = e i n= r n= X−∞ X−∞ This is a very remarkable formula. It gives a functional relation: the transformation τ 1/τ almost leaves the function unchanged; it changes only by a simple algebraic→− function. This is one of the achievements of Jacobi. In the earlier equations we can now put C(τ) = √(i/τ). In particular envis- V age 1′:

3 i 1 V ′(0/τ) = V ′ 0/ 1 τ 1 − τ  r  !   1  τ τ or V ′ 0/  =  V ′(o/τ) 1 − τ i · i 1 ! r But

V = πiτ/4 ∞ 2πimτ 3 1′(0/τ) 2πe (1 e ) = − Ym 1 πiτ/4 ∞ 2πim/τ 3 τ τ πiτ/4 ∞ 2πimτ 3 ∴ e− (1 e− ) = e (1 e ) = − i · i = − Ym 1 r Ym 1 12. Lecture 88

Extracting cube roots on both sides, 113

πiτ/12 ∞ 2πim/τ τ πiτ/12 ∞ 2πimτ e− (1 e− ) = ǫ e (1 e ) = − i = − Ym 1 r Ym 1 where ǫ3 = 1. Dedekind ﬁrst introduced the function

∞ η(τ) = eπiτ/12 (1 e2πimτ) = − Ym 1 Then 1 τ η = ǫ η(τ) − τ i ! r This is challenging; we have to decide which ǫ to take: ǫ3 = 1. The quotient 1 τ η( τ ) i η(τ) is an analytic (hence contains) function in the upper half-plane and− so must be situated in on of the three open sectors. Now make a special . choice;p put τ = i. Then η(i) = ǫ(+1)η(i), or ǫ = 1. 1 τ ∴ η = η(τ) − τ i ! r What we have done by considering the lattice of periods can be done in 114 more sophisticated ways. One can have a whole general theory of the transfor- aτ + b mations from 1, τ, to 1, . The quotients appear ﬁrst and can be carried cτ + d over. We start with V1 and come back to it; there may be diﬃculty, however in deciding the sign. Lecture 13

We arrived at the following result last time: 115 1 τ η = η(τ). − τ i ! r We began by investigating a transformation of V1(V , τ). Instead of looking upon 1 and τ as generators of the period lattice, we looked upon τ and 1 as generators (turning the plane around through arg τ): 1, τ τ, 1. We− have of course still the same parallelogram of periods. Since we→ should− like to keep 1 the ﬁrst period 1, we reduced everything by τ : τ, 1 1, τ ; so we had to V V V V V V 1 − → − investigate 1( τ/τ). 1( τ/τ) and 1( / τ ) have the same parallelogram of periods. − We could do this a little more generally. Let us introduce linear combinations: ω1 = cτ + d, ω2 = aτ + b, and go from ω1 to ω2 in the positive sense. In order that we must have these also as generating vectors for the same lattice, we should have a, b, c, d integers with

a b = 1. c d

Moreover we want the ﬁrst period to be always 1. (This is the diﬀerence

between our case and the Weierstrassian introduction of periods, where we have complete homogeneity). So replacing by linearity, the periods are 1 and = aτ+b τ′ cτ+d .

89 13. Lecture 90

Be sure that we want to go from 1 to τ′ through an angle less than π in the 116 positive sense. For this we want τ′ to have a positive imaginary parts Imτ¯′ > 0, or τ τ¯ ′ − ′ > 0 i 1 aτ + b aτ¯ + b i.e., > 0 i cτ¯ + d − cτ¯ + d ! 1 adτ + bcτ¯ adτ¯ bcτ i.e., − − > 0 i cτ + d 2 | | 1 (ad bc)(τ τ¯) i.e., − − > 0 i cτ + d 2 | | or since τ τ¯ is purely imaginary, − ad bc = 1. − ± We could do the same thing in all our different steps. The most important step, however, cannot be carried through, because we get lost at an important point; and rightly so, it becomes cumber some because a number-theoretic problem is involved there. Let us see what we have done. Compare aτ + b V ((cτ + d)V /τ) and V V 1 1 cτ + d . ! We want periods 1, τ′; indeed all things obtainable from ω1 = cτ + d and ω2 = aτ + b; or m1ω1 + m2ω2 must in their totality comprise all periods. For the first cτ + d is indeed a period, and for the second aτ + b. Now define 117 f (V ) = V1((cτ + d)V /τ) f (V + 1) is essentially f (V ):

f (V + 1) = V1((cτ + d)V + cτ + d/τ) c+d c2πiτ (cτ+d)V = ( ) e− e−∗∗∗∗∗ V ((cτ + d)V /τ), from the table, − 1 = ( ) f (V ) ······ f (V + τ′) = V1((cτ + d)ν + aτ + b/τ) a+b a2πiτ 2πia(cτ+d)ν = ( ) e− e− V ((cτ + d)V /τ) − 1 = ( ) f (V ) ······ f (V ) has, leaving trivial factors aside, periods 1, τ′ *****. So too for the V V aτ+b second function 1 cτ+d . . 13. Lecture 91

We can form quotients and proceed as we did earlier. Let us consider for a moment the V ′ s with double subscripts. This is a digression, but teaches us a good deal about how to work with V -functions. Recall that

2 νn πiτ(n+ 1 ) 2πiν(n+ 1 ) Vµν(ν/τ) = ( ) e 2 e 2 n − X V1(ν/τ) = Vn(ν/τ)

V2(ν/τ) = V10(ν/τ)

V3(ν/τ) = V00(ντ)

V4(ν/τ) = V01(ν/τ)

We take one liberty from now on. Take µ, ν to be arbitrary integers, no 118 longer 0, 1. That will not do very much harm either. In fact,

Vµ,ν+2(ν/τ) = Vµ,ν(v/τ) It is unfortunately not quite so easy for the other one:

ν Vµ+ ,ν(v/τ) = ( ) Vµ,ν(v/τ) 2 − For

ν πiτ(n+µ/2)2 2πiν(n+1+µ/2) Vµ+2,ν(v/τ) = ( ) e e n − X 2 = ( )ν( )v(n+1)eπiτ(n+1+µ/2) e2πiv(n+1+µ/2) n − − Xv = ( ) Vµν(v/τ), − on shifting the summation index from n to n + 1. The original table will be + + 1 + τ + 1+τ considerably reduced now; only in place of ν 1, ν 2 , ν 2 , ν 2 it + k + l will be now necessary to have the combination ν 2 2 τ. The expression k l for V ν + + τ/τ will include everything that we have done so far in one µν 2 2 single formula. !

k l νn πiτ(n+ µ )2 2πiν(n+ µ ) πi(k+lτ)(n+ µ ) V ν + + τ/τ = ( ) e 2 e 2 e 2 µν 2 2 n − ! X kµ (ν+k)n πiτ(n+µ/2+l/2)2 πiτl2/4 2πiν(n+µ/2+l/2) πilν = i ( ) e e− e e− n − X kµ πiτl2/4 πilν = i e− e− Vµ+l,ν+k(ν/τ)(*) 13. Lecture 92

119 This one formula has the whole table in it. We now turn to our purpose, viz. To consider the quotient

V1((cτ + d)ν/τ) V aτ+b 1 ν cτ+d We wish to discuss the behaviour a. little more explicitly of f (ν).

f (v) = V11((cτ + d)ν/τ)

f (v + 1) = V11((cτ + d)ν + cτ + d/τ)

f (v + τ′) = V11((cτ + d)ν + aτ + b/τ) putting k = 2c, l = 2d, µ = ν = 1 in (*), 120

d πiτc2 2πic(cτ+d)ν f (ν + 1) = ( ) e− e− V + , + ((cτ + d)ν/τ) − 1 2c 1 2d c+d πiτc2 πic(cτ+d)ν = ( ) e− e− f (v) − Similarly, putting k = 2a, l = 2b, µ = ν = 1,

a+b πiτa2 2πia(cτ+d)ν f (ν + τ′) = ( ) e− e− f (v). − Also deﬁning g(ν):

aτ + b V v = g(v) = V (ν/τ′), 1 cτ + d 11 . ! we have g(ν + 1) = V (ν + 1/τ′) = V (ν/τ′). 11 − 11 And putting k = 0, l = 2, µ = 3, ν = 1 in (*),

πiτ′ 2πiν g(ν + τ′) = e− e− V31(ν/τ′) πiτ 2πiν = e− ′ e− g(ν). − We form now in complete analogy with the old procedure f (ν) Φ(ν) = g(ν) c+d+1 πiτc2 2πic(cτ+d)νΦ(ν) Φ(ν + 1) = ( ) e− e− , − a+b+1 πiτc2 2πia(cτ+d)ν πiτ +2πiv Φ(ν + τ′) = ( ) e− e− e ′ Φ(ν) − 13. Lecture 93

Φ takes up exponential factors which contain ν linearly. As before ewe can 121 submerge this under a general form. Deﬁne

Ψ(ν) =Φ(ν)eh(ν),

where h(ν) is to be so determined that

Ψ(ν + 1) = Ψ(ν + τ′) = Ψ(ν)

we therefore want

h(ν+1) h(ν) c+d+1 c2πiτ 2πic(cτ+d)ν e − ( ) e− − = 1, − h(ν+τ ) h(ν) a+b+1 a2πiτ+πiτ +2πiν 2πia(cτ+d)ν e ′ − ( ) e− ′ e− = 1. − It will be convenient to observe that c + d + cd + 1 = (c + 1)(d + 1) is even, for at least one of c, d should be odd as otherwise c, d would not be co-prime and we would not have a b = 1 c d

So ( )c+d+1 = ( )cd = eπicd. h is given by the equations: − − h(ν + 1) h(ν) = 2πic(cτ + d)ν + πic(cτ + d), − h(ν + τ′) h(ν) = 2πia(cτ + d) + πia(aτ + b) πiτ′ − − = 2πc(aτ + b)ν + πicτ′(aτ + b).

We have to introduce a suitable function h(ν). Since the diﬀerence equation can be solved by means of a second degree polynomial, put

h(ν) = Aν2 + B for each separately and see whether it works for both. 122

h(ν + δ) h(ν) = 2Aνδ + Aδ2 + Bδ − = δ(2Aν + Aδ + B)

Putting δ = 1, τ′, we ﬁnd that A = πic(cτ + d) works in both cases. Also for δ = 1,

A + B = πic(cτ + d), aτ + b 2 aτ + b πic(aτ + b)2 A + B = cτ + d cτ + d cτ + d ! ! 13. Lecture 94

So B = 0 ﬁts both. Hence

h(ν) = Aν2, A = πi(cτ + d)c

2 f (ν) ∴ Ψ(ν) = eπic(cτ+d)ν g(ν) And this is a doubly periodic entire function (because the numerator and denominator have the same simple zeros) and therefore a constant. We thus have the transformation formula

aτ + b 2 V ν = Ceπic(cτ+d)ν V ((cτ + d)ν/τ) 11 cτ + d 11 . ! where C may depend on the parameters τ, a, b, c, d:

C = C(τ; a, b, c, d)

More generally we can have a parallel formula for any µ, ν. As before we 123 get an equation for C2. And there the thing stops. Formerly we were in a very good position with the special matrix

a b 0 1 = − . c d 1 0          For general a, b, c, d we get into trouble.  Lecture 14

We were considering the behaviour of V11(ν/τ) under the general modular 124 transformation:

aτ + b 2 V ν = C(τ)eπiC(cτ+d)ν V ((cτ + d)ν/τ), (1) 11 cτ + d 11 . ! a b a, b, c, d integers with =+1. c d

We want to determine C (τ) as far as possible. We shall do this up to a

sign. ν is unimportant at the moment; even if we put ν = 0, C(τ) survives. Put± + = 1 τ′ 1 τ′ ν 2 , 2 , 2 in succession, ans use out auxiliary formula which contracted the whole table into one thing:

k lτ kµ πiτl2/4 πiν V v + + τ = i e− e− V + + (v/τ)(*) µν 2 2 µ l,v k . ! aτ + b Putting ν = 1 in (1), and writing τ = , 2 ′ cτ + d + 1 πic(cτ+d)/4 cτ d V τ′ = C(τ)e V τ (2) 11 2 11 2 . ! . ! This is the right moment to call for formula (*). From (*) with ν = 0, µ = ν = 1, k = 1, l = 0, we get 1 V τ′ = iV (0/τ′) 11 2 12 . ! Also from (*) with ν = 0, µ = ν = 1, k = d, l = C, we get + cτ d d πic2/4 V τ = i C− V + + (o/τ). 11 2 1 c,1 d . ! 95 14. Lecture 96

Substituting these two formulas in the left and right sides of (2) respec- 125 tively, we get

πic(cτ+d)/4 d πiτc2/4 iV12(0/τ′) = C(τ)e i e− V1+c,1+d(0/τ)

Now, recalling that

Vµ,ν+2(ν/τ) = Vµν(ν/τ) ν (**) Vµ+ ,ν(ν/τ) = ( ) Vµν(ν/τ), 2 − the last formula becomes

πicd/4 d iV10(0/τ′) = C(τ)e i V1+c,1+d(0, τ) (3)

Putting ν = τ′/2 in (1), we have + τ′ πic/4τ (aτ+b) aτ b V τ′ = C(τ)e ′ V τ . 11 2 11 2 . ! . ! Making use of (*) in succession on the left and right sides (with proper choice of indices) as we did before, this gives

2 πiτ′/4 πicτ′(aτ+b)/4ibe πia τ/4 e− V12(0/τ′) = C(τ)e − V1+a,1+b(0/τ), and this, after slight simpliﬁcation of the exponents on the right sides, gives in view of (**), b πiab/4 V (0/τ′) = C(τ)i e V + , + (0/τ) (4) − 01 1 a 1 b Putting ν = (1 + τ′)/2 in (1), 126 + + + + 1 τ′ πic/4(1+τ )(la+c)τ+b+d) (a c)τ l d V τ′ = C(τ)e ′ V τ 11 2 11 2 . ! . ! Again using (*) and (**) as we did earlier, this gives

+ + + + + + 2 πiτ′/4V = πic/4(1 τ′)((a c)τ l d) l d πi(0 c) /4V (0/τ) ie− 22(0/τ′) C(τ)e i e− 1+a+c,1+l+d This of course can be embellished a little:

+ + + + + + + 2 V = πi/4(1 τ′)(c(a c)τ cb id 1) b d πi/4 πiτ(a c) /4V (0/τ) i 00(0/τ′) C(τ)e i e− e− 1+a+c,1+b+d πi/4(a+c)((a+c)τ+b+d) l+d πi/4 πiτ(a+c)2/4 = C(τ)e i e− e− V1+a+c,1+b+d V = πi/4(a+c)(b+d) b+d πi/4V (0/τ) ∴ i 00(0/τ′) C(τ)e i e− 1+a+c,1+b+d (5) 14. Lecture 97

Now utilise the formula: V = V V V 1′(0/τ) π 2(0/τ) 3(0/τ) 4(0/τ) Multiplying (3), (4) and (5),

3 b+d πi/4(ab+cd+(a+b)(b+d) 1) V ′ (0/τ′) = (C(τ)) ( ) e − 11 − iπV + , + (0/τ)V + , ++ (0/τ)V + + , + + (0/τ) × 1 c 1 d 1 a 1 b 1 a c 1 b d Observethat the sum of the first subscriptson the right side = 3+2a+2c 1 127 (mod 2). So either all three numbers 1 + a,1 + c,1 + a + c are odd, or one≡ of them is odd and two even. Then first case is impossible since we should then a b , have both a and c even and so c d 1. So two of them are even and one odd. The even suffixes can be reduced to zero and the odd one to 1 by repeated application of (**). Similarly for the second suffixes. So the V -factors on the right will be V00, V01, V10. What we hate is the combination 1, 1 and this does not occur. (If it did occur we should have V11 which vanishes at the origin). Although we can not identify the V -factors on the right, we are sure that we get exactly the combinations that are desirable: 01, 10, 00. The dangerous combination is just out. Let us reduce the subscripts by stages to 0 or 1 as the case may be. When we reduce the second subscript nothing happens, whereas when we reduce index by steps of 2, each time a factor 1 is introduced, by virtue of (**). By ± 1+c the time the subscript 1 + c is reduced to 0 or 1, a factor ( )[ 2 ](1 + d) will − have accumulated in the case of V1+c,1+d. Similarly in the case of V1+a,1+b and V1+a+c,1+b+d. Altogether therefore we have a factor 128

1+c (1+d)+ 1+a (1+b)+ 1+a+c (1+b+d) ( )[ 2 ] [ 2 ] [ 2 ] , − V V V and when this compensating factor is introduced we can write 00, 11′ and 10. Hence our formula becomes

3 α πi/4(ab+cd+(a+c)(b+d) 1) V ′ (0/τ) = (C(τ)) ( ) e − iπV (0/τ)V (0/τ)V (0/τ) (6) 11 − 00 01 10 where 1 + c 1 + a 1 + a + c α = b + d + (1 + d) + (1 + b) + (1 + b + d) 2 2 2 " # " # " # From (1), diﬀerentiating and putting ν = 0, we have V = + V 11′ (0/τ′) C(τ)(Cτ d) 11′ (0/τ) (7) 14. Lecture 98

Dividing (6) by (7)

2 α πi/4(ab+cd+(a+c)(b+d) 1) (C(τ)) = (cτ + d)( ) e− − − + cτ d α πi/4(ab+cd+(a+c)(b+d) 3) = ( ) e− − i − (we may assume c > 0, since c = 0 implies ad = 1 or a, d = 1, which give just translations). ±

+ cτ d α πi/8(ab+cd+(a+c)(b+d)1 3) ∴ C(τ) = i e− − ± r i For the square root we take the principal branch. Since Im(cτ + d) > 0, cτ + d cτ + d R > 0, so that is a point in the right half-plane. The sign is still i i uncertain. πi/8( ) The factor e− ··· lookslike a 16th root of unity, but is really not so. Since 129 ad + bc has the same parity as ad bc = 1, the exponent is even, and therefore what we really have is only an 8th− root of unity. What could we do now? We really do not know of any fruitful way. We cannot copy what we did formerly. There we had a very special case: τ′ = a b = 0 1 1/τ, or the modular substitution involved was c d 1− 0 . The sign depends− only on a, b, c, d, not on τ, so that it is enough if we make a special± choice of τ in the equation. Formerly we could take τ = τ′ = i and it worked so beautifully because τ is a study the fixed points of the transformation τ′ = aτ + b . The fixed points ξ are given by cτ + d aξ + b ξ = , cξ + d or cξ2 + (d a)ξ b = 0 − − (a d) (a d2) + 4bc i.e., ξ = − ± − p2c (a d) (a + d)2 4 = − ± − p2 since ad bc = 1. Hence− we have several possibilities. If the square root is imaginary we have twp points one in each of the upper and lower half-planes, and for this a + d < 130 2, so that the square root becomes √ 4 or √ 3 according as a + d =| 0or| 1. This is the elliptic case. If a + d = 2,− we have− one rational fixed| point;| this is the parabolic case. And in| the huge| infinity of cases, a + d > 2, we have two | | 14. Lecture 99

real fixed points -the hyperbolic case. Here the fixed are not accessible to us because they are quadratic algebraic numbers on the real axis. In the elliptic case with a + d < 2 we could finish the thing without much trouble. In the parabolic case| we| are already in a fix. Much more difficult is the hyperbolic case.

If ξ1 and ξ2 are the ﬁxed points, τ and τ′ will lie on the same circle through ξ1 and ξ2, and repetitions of the transformation would give a sequence of points on the same circle which may converge to either ξ1 or ξ2. So the ambiguity in the sign will remain. ± It will be much more diﬃcult when we pass from V to η, because then we shall have to determine a cube root. Lecture 15

We were discussing the possibility of getting a root of unity determined for 131 aτ + b the transformation of V ν . There do exist methods for determining 11′ cτ + d this explicitly. Only we tried . to carry! out the analogue with the special case as far as possible, not with complete success. other methods exist. The first of these is due to Hermite, done nearly 100 years ago. He used what are called Gaussian sums. There are difficulties there too and we want to avoid them. Another method is that of Dedekind using Dedekind sums. = 1 In the special case of the transformation from τ to τ′ τ we were faced a b − with an elliptic substitution c d . These are of two sorts: 1. a + d = 0 2. a + d = 1 ± In both cases we can completely forget about the root of unity if we remember the following fact. Our formula had the following shape: aτ + b cτ + d cτ + d V = ρ , , , V /τ 11′ 0 + (a b c d) 11′ (0 )(*) cτ d r i · i . ! where ρ is a root of unity which is completely free of τ. we can then get things straightened out. We have only to consider the fixed points of the transformation given by a d (a + d)2 4 ξ = − ± − p2c aξ + b Put ξ on both sides of the formula; since ξ = , both sides look alike 132 cξ + d V and 11′ does not vanish for appropriate τ in the upper half-plane (we may take c > 0), so that ρ is given directly by the formula.

100 15. Lecture 101

Case 1. a + d = 0 2d √ d d + i ξ = − ± − = − 2c c (reject the negative sign since we want a point in the upper half-plane). cξ + d i = = i i 1 V = V ∴ 11′ (0/ξ) ρ(a, b, c, d) 11′ (0/ξ) So ρ(a, b, c, d) = 1 and remains 1 in the general formula when we go away from ξ. Case 2. a + d = 1 ± 1 2d + √ 3 ξ = ± − − 2c 1 + i √3 ∴ cξ + d = ± = eπi/3 or e2πi/3 2 + cξ d πi/3 πi/2 2πi/3 πi/2 = e − or e − i πi/6 = e± + cξ d πi/12 = e∓ r i Putting ξ on both sides of (*), 133

πi/4 1 = e∓ ρ(a, b, c, d)

πi/4 ∴ ρ(a, b, c, d) = e± when a + d = 1, (we may take c > 0; the case c = 0 is uninteresting and if c < 0 we can make± it c > 0). There are unfortunately no more cases like these. Parabolic case. The analysis here is a little longer but it is worth while working it out. Now a + d = 2, and there is only one fixed point ± a d 2d 2 d 1 δ ξ = − − ± = − ± = 2c − 2c c −γ where (γ,δ) = 1 and we may choose γ > 0. The fixed point is now a rational point on the real axis. We try to approach it. This is a little difficult because 15. Lecture 102

we do not know what the function will do thee. But by an auxiliary transformation we can throw this point into the point at inﬁnity. Consider the auxiliary transformation Aτ + B A B T = , = 1 γτ + δ γ δ

The denominator becomes zero for τ = ξ. Let

+ = Aτ′ B T ′ + γτ′ δ (notice that γ and δ have got something to do with the properties of two other 134 numbers c, d). Now (*) gives

+ 3 V = γτ δ V 11′ (0/T) ρ(A, B,γ,δ) 11′ (0/τ),  r i    3  γτ + δ V =  ′  V 11′ (0/T ′) ρ(A, B,γ,δ) 11′ (0/τ′).  r i    Dividing, we get   3 + γτ′ δ V ′ (0/T ′) i V ′ (0/τ′) 11 = 11 (1) V  q +  V 11′ (0/T) γτ δ 11′ (0/τ)  i     q  The left side gives the behaviour at inﬁnity. We cannot of course put τ = ξ. Put τ = ξ + it, t > 0, and later make t 0. τ is a point in the upper half-plane. → τ ξ = it, − aτ + b aξ + b τ′ ξ = − cτ + d − cξ + d τ ξ = − (cτ + d)(cξ + d) = it 1 ict ± This is also in the upper half-plane. τ ξ as t 0 ′ → → Let us calculate T and T ′. For this consider

+ + = Aτ′ B Aτ B T ′ T + + − γτ′ δ − γτ δ 15. Lecture 103

= τ′ τ + − + (γτ′ δ)(γτ δ) = c ∓γ2 135 This is quite nice; the diﬀerence is a real number. Aτ + B A(ξ + it) + B T = = γτ + δ γ(ξ + it) + δ Ait + Aξ + B = , since γξ + δ = 0, γit A δ + B = A + − γ γ γit A Bγ Aδ = + − γ γτt A 1 = + , since Bγ Aδ = 1 γ γ2t − −

i (along the ordinate x = A ) as t 0 → ∞ γ → A c i T ′ = + γ ± γ2 r2t i as t 0 → ∞ → Now recall the inﬁnite product formula 136

V = πiT/4 ∞ 2πinT 3 11′ (0/T) 2πie (1 e ) = − Yn 1 = A + i Let T γ γ2t . Then

2πinT 2πinA/γ 2πnt/γ2 e = e e− 0 → V = πiT/4 ∴ 11′ (0/T) 2πie (a factor tending to 1) We do not know what happens to eπiT/4. But we need only the quotient. So

V ′ (0/τ) 11 πi(T ′ T)/4 V e − 11′ (0/τ) ∼ πic/4γ2 = e∓ (2) 15. Lecture 104

V V + = Consider similarly the quotient 11′ (0/τ′)/ 11′ (0/τ). We have, since γξ δ 0, γτ + δ γ(ξ + it) + δ = = γt i i γτ + δ or = √νt, where we take the positive square root i r + it + γτ′ + δ γ ξ 1 ict δ γt = ± = i i 1 ict ± γτ + δ 1 ∴ ′ = √rt (both branches principal) i 1 ict r r ± √γt as t 0. ∼ →

γτ′ + δ γτ+δ Hence the quotient behaves like 1. 137 i i r And so we have what we were. after:q V 11′ (0/τ′) πic/4γ2 V e∓ as t 0 (3) 11′ (0/τ) → → + + + a d + cit + d cτ d = c(ξ it) d = −2 i i i a + d + cit + = 2 = 1 cit i ± i = i + ct ∓ i as t 0. →∓ → What will the square root of this do? cτ + d = √ct i, and this i ∓ doesr lie in the proper half plane because ct > 0. For small t it will be very near the imaginary axis near i. So the square root lies in the∓ sector, in the lower half plane if we choose πi/4 √ i = e− , and in the upper half-plane− if we choose √+i = cτ + d eπi/4 Hence e πi/4 i → ∓ as t 0. r → 15. Lecture 105

Using this fact as well as (2) and (3) in (*) we get 138

πic/4γ2 3πi/4 e∓ = e∓ ρ(a, b, c, d)

πi 2 ∴ ρ(a, b, c, d) = e∓ (c/γ 3) 4 − We observe that the common denominator (γ,δ) = 1 plays a role, however a and b do not enter. Hyperbolic case. The thing could also be partly considered in the hyperbolic case. It will take us into deeper things like real quadratic fields and we do not propose to do it. Letus returnto whatwe had achievedin the specificcase. We had a formula for η(τ): cτ + d η(τ′) = ǫ(a, b, c, d)η(τ), r i where ǫ(a, b, c, d) is a 24th root of unity. This shape we have in all circumstances. The difficulty is only to compute ǫ. We shall not determine it in general, and we can do away with it even for the purpose of partitions by using a method developed recently by Selberg. However in each specific case we can compute ǫ.

1 τ η = η(τ) − τ i ! r Now

∞ η(τ) = eπiτ/12 (1 e2πinτ), = − Yn 1 η(τ + b) = eπib/12η(τ)

Out of these two facts we can get every other one, because the two substi- 139 tutions 1 1 0 1 S = , T = − 0 1 1 0          form generators of the full modular  group. This can be shown as follows. Take c > 0.

aτ + b = q (cτ + d) a, τ b , c > a , 0 − − 1 | 1| 15. Lecture 106

+ + aτ d = a1τ b1 c d = or + q0 + , 1 cτ d − cτ d a b 1 1

(if a < 0 this step is unnecessary). Similarly

cτ + d a2τ + b2 = q1 a1τ + b1 − a1τ + b1 We thus get a continued fraction expansion. The partial quotients get simi- τ + b lar and simpler every time and end with = τ + q . so wecan goback and 0 + 1 k take linear combinations; all that we have to do is either to add an integer to τ or take 1/τ. As an− example, let us consider

3τ + 4 η 2τ + 3 ! 3τ + 4 Let us break into simpler substitutions, 2τ + 3 3τ + 4 1 τ3 = = 1 , 2τ + 3 − τ2 τ = 2 + τ ; 2 − 1 1 τ = 1 −τ + 1 1 τ + 1 η(τ ) = η = η(τ + 1) 1 −τ + 1 i ! r τ + 1 = eπi/12η(τ). r i πi/6 η(τ ) = η(τ 2) = e− η(τ ) 2 1 − 1 + τ 1 πi/12 = e− η(τ) r i · 1 τ η = 2 η(τ ) −τ i 2 2 ! r + τ 1 τ2 πi/12 = e− η(τ) i i r r The two square roots taken separately are each a principal branch, but taken 140 15. Lecture 107

together they may exceed one. We can write this as

+ 1 1 τ 1 2 τ+1 πi/12 η = − − e− η(τ) −τ i s i 2 ! r + = τ 1 2τ 3 πi/12 − +− e− η(τ) r i s i(τ 1)

2τ 3 πi/12 = − − e− η(τ) r 1 − πi/12 = √3 + 2τe− η(τ) ± Here we are faced with a question: which square root are we to take? 141 √ + = πi/4 2τ+3 We write 3 2τ e i Let us look into each rootq singly. For τ = it where do they go? + + τ 1 = it 1 r i r i with argument 0 as t . →∞ →∞ 2τ 3 2it 3 − − = − − √2i as t , s i(τ + 1) s i(τ + 1) → →∞ or its argument = π/4

τ + 1 2τ 3 The product − − has here argument π/4, so that it continues i i(τ + 1) to be the principalr branch.r Of course in a less favourable case, if we had two other arguments, together they would have run into something which was no longer a principal branch. Finally,

3τ + 4 2τ + 3 η = eπi/4 η(τ) 2τ + 3 i ! r and here there is no ambiguity. Actually in every speciﬁc case that occurs one can compute step and make sure what happens. There does exist a complete formula which determines ǫ(a, b, c, d) explicitly by means of Dedekind sums S (h, k). Part III

Analytic theory of partitions

108 Lecture 16

Our aim will be now to get an explicit formula for p(n) and things connected 142 with it. Later we shall return to the function η(τ) and the discussion of the sign of the square root. That will again lead us into some aspects of the theory of V -functions connected with quadratic residues. Let us come to our topic. Euler had, as we know, the identity:

∞ 1 p(n)xn = . n=0 ∞ (1 xm) X m=1 − Q This is the starting point of the function-theoretic treatment of p(n). 1 f (x) p(n) = dx, 2πi xn+1 Zc

∞ m 1 where f (x) = (1 x )− and C is a suitable closed path contained in the m=1 − unit circle, in whichQ the function is analytic, and enclosing the origin. Since p(n)xn is a power series beginning with 1, this means a little more. n may be n 1 negative also; and when n is negative f (x)x− − is regular at x = 0. Therefore weP include negative exponents also in our discussion; we put p( n) = 0, n > 0, when is convenient. Hereafter we shall take n to be an integer−R 0; we shall choose n and keep it ﬁxed throughout our discussion. It is a little more comfortable to change the variable and put x = e2πiτ, 143 Im τ> 0, which is familiar to us. dx = e2πiτ 2πidτ and the whole thing boils down to · α+1 2πiτ 2πinτ p(n) = f (e )e− dτ Zα 109 16. Lecture 110

It is enough to take the integral along a path from an arbitrary point α to the point α + 1, because the integrand is periodic, with period 1. (This path replaces the original path C that we had in the x-plane before we changed the variable). The method of Hardy and Ramanujan was to take a curve rather close to the unit circle which is a natural boundary for the function (this will come out in the course of the argument). They cut up the path of integration into pieces called Farey arcs, and the trick was to replace the function by a simpler approximating function on each specific Farey arc. We shall use not exactly this method, but consider a special path from α to α+1, which we shall discuss. We shall keep our formula in abeyance for a moment and give a short discussion of Farey series (‘series’ here is not to be understood in the sense of infinite series, but as just an aggregate of numbers). Cauchy did make all the 144 observation attributed by Hardy and Wright to Farey; Farey made his remarks in the Philosophical Magazine, 1816. He put only questions; Cauchy had all the answers earlier. We deal with the interval (0, 1). Choose all reduced fractions whose denominators do not exceed 1, 2, 3, in succession. Let us write down the first few, with the fractions arranged in··· increasing order of magnitude.

0 1 1 1 order 1 0 1 1 1 2 1 order 2 0 1 1 2 1 1 3 2 3 1 order 3 0 1 1 1 2 3 1 1 4 3 2 3 4 1 order 4 0 1 1 1 2 1 3 2 3 4 1 1 5 4 3 5 2 5 3 4 5 1 order 5 The interesting fact is that we can write down a new in the following way. h h We repeat the old row and introducesome new fractions. If < ′ are adjacent k k fractions in a row, the new one introduced between these in the′ next row is 16. Lecture 111

h + h′ + , provided that the denominator is of the proper size. Following Hardy k k′ h + h h h h and Littlewood we call ′ the ‘mediant’ between and ′ . We have < k + k k k k h + h h ′ ′ ′ < ′ , so that the order is automatically the natural order. We call k + k k that row′ which′ has denominator k N, the Farey series of order N. We get this by forming mediants from the≤ preceding row. Farey made the following observation. Take two adjacent fractions in a row; then the determinant formed by their numerators and denominators is equal to 1. For instance, in the ﬁfth − 1 2 1 2 row and are adjacent and = 1***********. If we now prove that 3 5 3 5 −

new fractions are always obtained by using mediants, then we can be sure, by 145

induction, that this determinant is always 1. For, let − h h ′ = 1; then k k − ′

h h + h h + h h ′ = 1 = ′ ′ k k + k − k + k k ′ ′ ′

If indeed only mediants occur, Farey’s observation is justified. And this is so. Observe that these fractions must all appear in their lowest terms; otherwise, the common factor will show up and the determinant would not be 1. Suppose that we want to find out where a particular fraction appears. Say,− we H have in mind a specific fraction . It should occur for the first time in the K Farey series of order N = K and it should not be present on any series of order H < K. Now look at N = K 1 where is not present. If we put it in, it will − K h h belong somewhere according to its size, i.e., we can find fractions 1 , 2 , with k1 k2 h1 H h2 k, k2 < N such that < < . Assume that the determinant property and k1 K k2 the mediant property are true for order N < K. (They are clearly true up to order 5, as we verify by inspection, so that we can start induction). Now prove h them for N = K. Try to determine H and K by interpolation between 1 and k1 h 2 . Put k2 Hk Kh = λ, 1 − 1 16. Lecture 112

Hk + Kh = µ, − 2 2 so that λ and µ are integers > 0. Solving for H and K by Cramer’s rule, 146

λ h λ k − 1 − 1

µ h2 µ k2 H = , K =

h k h k 2 2 2 2 h k h k 1 1 1 1

= By induction hypothesis, the denominator 1, and so

H = λh2 + µh1

K = λk2 + µk1 H λh + µh or = 2 1 . K λk2 + µk1 What do we know about K? K did not appear in a series of order K 1; − k1 and k2 are clearly less than K. What we have found out so far is that any h h λh + µh fraction lying between 1 and 2 can be put in the form 2 1 . Of these k1 k2 λk2 + µk1 only one interests us - that one with lowest denominator. This comes after the ones used so far. Look for the one with lowest denominator; this corresponds to the smallest possible λ, µ, i.e., λ = µ = 1. Hence ﬁrst among the many later H h + h appearing ones is = 1 2 , i.e., if in the next Farey series a new fraction K k1 + k2 is called for, that is produced by a mediant. So what was true for K 1 is true for K; and the thing runs on. − One remark is interesting, which was used in the Hardy - Littlewood- Ra- h h manujan discussion. In the Farey series of order N, let 1 and 2 be adjacent k1 k2 h h h + h fractions. 1 < 2 1 2 does not being these. It is of higher order. This k1 k2 · k1 + k2 says that k1 + k2 > N. For two adjacent fractions in the Farey series of order N, 147 the sum of the denominators exceeds N. Both k and k N, so 1 2 ≤ 2N k + k > N. ≥ 1 2

k1 and k2 are equal only in the ﬁrst row, otherwise it would ruin the determinant rule. So 2N > k1 + k2 > N, N > 1. 16. Lecture 113

This was very often used in the Hardy - Ramanujan discussion. (The Farey series is an interesting way to start number theory with. We can derive from it Euclid’s lemma of decomposition of an integer into primes. This is a concrete way of doing elementary number theory). We now come to the special path of integration. For this we use Ford Circles (L.R. Ford, American Mathematical Monthly, 45 (1938), 568-601). We describe a series of circles in the upper half-plane. To each proper fraction h h i 1 we associate a circle C with centre τ = + and radius , so the k hk hk k 2k2 2k2 circles all touch the real axis.

Take another Ford circle Ch′k′ , with centre at τh′k′ . Calculate the distance between the centres.

h h 2 1 1 2 τ τ 2 = ′ + . | hk − h′k′ | k − k 2k2 − 2k 2 ′ ! ′ ! = 1 + 1 The sum of the radii 2 2 148 2k 2k′ 1 1 h h 2 1 τ τ 2 + = ′ | hk − h′k′ | − 2k2 2k 2 k − k − k2h 2 ′ ! ′ ! ′ (hk h k)2 1 = ′ ′ , − 2 2 − 0 k k′ ≥ since h, k are coprime and so h k is an integer , 0. So two Ford circles never h′ k′ intersect. And they touch if and only if

h k = 1, ± h′ k′

h h′ i.e., if in a Farey series , have appeared as adjacent fractions. k k′ 16. Lecture 114

Now we come to the description of the path of integration from α to α + 1. For this consider the Ford circle Chk.

h h In a certain Farey series of order N we have adjacent fractions 1 < < k1 k h2 . (We know exactly which are adjacent ones in a speciﬁc series). Draw also 149 k2

the Ford circles Ch1k1 and Ch2k2 . These touch Chk. Take the arc γhk of Chk from one point of contact to the other in the clockwise sense (the arc chosen is the one not touching the real axis). This we do for all Farey fractions of a given order. We call the path belongingto Farey series of order NPN . Let us describe this in a few cases. We ﬁx α = i and pass to α + 1 = i + 1. Take N = 1; we have two circles of i i radii 2 each with centres at and 1 + 2 2

1 + i 1 + i + ρ1 will be the path consisting of arcs from i to 2 2 and 2 2 to i 1. Later because of the periodicity of f (e2πiτ) we shall replace the second piece by the 16. Lecture 115

0 1 arc from 1 + i to i. Next consider Farey series of order 2; and are no − 2 2 1 1 longer adjacent. The path now comprises the arc of C01 from i to the point of contact with C12, the arc of C12 from this point to the point of contact with C11 150 and the arc of C11 from this point to i + 1. Similarly at the next stage we pass from i on C01 to i+1on C11 through the appropriate arcs on the circles C13, C12, C23 in order. So the old arcs are always retained but get extended and new arcs spring into being and the path gets longer and longer. At no stage does the path intersect itself, but these are points of contact. The path is complicated and was not invented in one sitting. The Farey dissection of Hardy and Ramanujan can be pictured as composed of segments parallel to the imaginary axis. Here it is more complicated. We need a few things for our consideration. We want the point of contact

of Chk and Ch′k′ . This is easily seen to be the point

1 1 2 2 2 2 h i k h′ i k′ τ 2k′ + τ 2k = + + + hk 1 + 1 h′k′ 1 + 1 2 2 + 2 2 2 + 2 2 2 2 k 2k k k′ k′ 2k′ k k′ 2k 2k′ 2k 2k′2 ! ! 2 = h + h′ h k′ + i k k − k k2 + k 2 k2 + k 2 ′ ! ′ ′

and this, since 151

h h h h h k i < ′ and ′ = 1, is = + ′ + k k k k(k2 + k 2) k2 + k′2 ′ k′ k ′

h = + ξ′ k hk 16. Lecture 116

k i where ζ = ′ + . We notice that the imaginarypart1/(k2+k 2) hk′ 2 + 2 2 + 2 ′ k(k k′ ) k k′ gets smaller and smaller as k + h′ lies between N and 2N. Each arc runs from h h + ζ to + ζ . Such an arc is the arc ν . No arc touches the real axis. k hk′ k hk′′ hk We continue our study of the integral. Choose a number N, the order of the Farey series, and cut the path of integration PN into pieces γhk.

2πiτ 2πinτ p(n) = f (e )e− dτ

PZN

2πiτ 2πinτ = f (e )e− dτ = (h,k) 1 γZ 0 Xh

ζhk′′ 2πi( h +ζ) 2πin( h +S ) p(n) = f (e k )e− k dS = (h,k) 1 ζZ 0 Xh

zhk′′ 2πinh/k z e− 2πi( h + i ) 2πnz/k2 = k k2 z p(n) i 2 f (e )e d = k (h,k) 1 zZ 0 Xh

Now ﬁnd out zhk′ and zhk′′ .

k2 + ikk z = ′ hk′ 2 + 2 k k′ 2 k ikk′′ z′′ = − hk 2 + 2 k k′′ So what we have achieved so for is to cut the integral into pieces. 153 16. Lecture 117

The whole thing lies on the right half-plane. The original point of contact is 0 and everything 1 1 lies on the circle z = . | − 2| 2 This is a normalisation. We now study the complicated function ǫ be- on each arc separately. We shall 0 ﬁnd that it is practically the function η(τ) about which we know a good deal:

+ + aτ b = cτ d η + ǫ η(τ), cτ d r i ing a complicated! 24th root of unity. Lecture 17

We continue our discussion of p(n). Last time we obtained 154

zhk′′ i 2πinh/k 2πi( h + z ) 2πinz/k2 = k k2 z p(n) 2 e− f e e d = k (h,k) 1 zZ 0 Xh

2πiτ πiτ/12 1 f (e ) = e (η(τ))− , 1 since f (x) = , ∞ (1 xm) m=1 −

∞Q η(τ) = eπiτ/12 (1 e2πimτ). = − Ym 1 h iz For us τ = + . k k2 We can now use the modulartransformation. We want to make Im τ large so aτ + b that we obtain a big negative exponent. So we put τ = , a, b, c, d being ′ cτ + d chosen in such a way for small τ, τ′ becomes large. This is accomplished by 155 taking kτ h in the denominator; kτ h = 0 when z = 0 and close to zero − − aτ + b when z is close to the real axis. We can therefore put τ = where a, b ′ kτ h a b = =− should be integers such that k h 1. This gives ah bk 1 or ah 1 − − − ≡ −

118 17. Lecture 119

(mod k). Take a solution of this congruence, say h′ i.e., choose h′ in such a way that h′h 1 (mod k), which is feasible since (h, k) = 1. As soon as h′ is found, we can≡− ﬁnd b. Thus the matrix of the transformation would be

+ a b h hh′ 1 = ′ − k c d  k h     −      So we have found a suitable τ′for our purpose.

h iz hh′ + 1 h′ + k k2 − k τ′ = ! h iz k + h k k2 − ! iz h′ 1 = k − iz h i = ′ + . k z If z is small this is big. Now recall the transformation formula for η: if c > 0,

aτ + b cτ + i η = ǫ η(τ) cτ + d i ! r In our case 156

2πiτ πiτ /12 1 f (e ′ ) = e ′ (η(τ′))− + 1/2 πiτ /12 1 cτ d − 1 = e ′ ǫ− (η(τ))− i ! + 1/2 πiτ /12 1 cτ d − πiτ/12 2πiτ e ′ ǫ− e f (e ) i ! And this is what we were after. Since h iz iz cτ + d = kτ h = k + h = , − k k2 − k ! this can be rewritten in the form:

2πiτ 2πi h + iz f (e ) = f (e k k2 ) 17. Lecture 120

πi πi i h′ i h h iz 2πi + ′ 2 z k − k k − z  k z  = ǫe12 e12   f e          k     r        and there is no doubtabout the square root - it is the principa l branch. We write

πi h h′ = 12  k − k  ωhk ǫe       So something mathematical has happened after all this long preparation; and we can make some use of our previous knowledge. We have

zhk′′ z iωhk 2πinh/k π 1 2πi h′ + i 2πnz/k2 = ′ 12 z − k2 z k z z p(n) 5/2 e− e √ f e e d k o h

π 1 z 2 Ψk(z) = √ze 2 z − k Then

zhk′′ iωhk 2πinh/k 2πnz/k2 p(n) = ′ e− e Ψ (z)dz+ k5/2 k 0 h

zhk′′ iωhk 2πinh/k 2πi h′ + i 2πnz/k2 Ψ z k z z 5/2 e− k( ) f (e ) 1 e d k − o h

to use an estimate for the small contribution from the second term. We have

now to appraise this. Writing Ihk and Ihk∗ for the two integrals, we have

iωhk 2πinh/k iωhk 2πinh/k p(n) = ′ e− I + ′ e− I∗ k5/2 hk k5/2 hk o h

Here we stop for a moment and consider only Ihk∗ and see what great advantage we got from our special path.

This is the arc of the circle z 1 = 1 from z to z described clockwise. | − 2 | 2 hk′ hk′′ = ∞ ν Since f (x) 1 p(ν)x , the integrand in Ihk∗ is regular, and so for integration − m=1 z z we can just as wellP run across, along the chord from hk′ to hk′′ . Let us see what happens on the chord. We have

2 π πz + 2πnz 2πih′ /k 2πz 2πnz/k 2πih′ /k 2π/z f (e − ) 1 Ψ (z)e = f (e − ) 1 √ze 12z − 12k2 k2 − k − R π Rz π ( 1 +2n) ∞ (2πi h′ 2π )ν = √ze 12z e k2 − 2 p(ν)e k − z × = Xν 1

∞ R 1 (2πν π ) π ( 1 +2n)Rz √z p(ν)e− z − 12 e k2 − 12 ≤| | = Xν 1 Let us determine Rz and R 1 on the path of integration o < Rz 1 on the 159 z ≤ R 1 chord. And z > 1; for 1 1 x R = R = , z x + iy x2 + y2 17. Lecture 122

while the equation to the circle is (x 1 )2 + y2 = 1 or x2 + y2 = x; the interior − 2 4 of the circle is x2 + y2 < x, and so R 1 1, equality on the circle. z ≤ √z the longer of the distances of z , z from 0. | |≤ hk′ hk′′ We have already computed zhk′ and zhk′′ : 160

2 k kk1 z′ = + i , hk 2 + 2 2 + 2 k1 k k1 k 2 k kk2 z′′ = + i hk 2 + 2 2 + 2 k2 k k2 k 4 + 2 2 2 2 k k k1 k z′ = = hk 2 + 2 2 2 + 2 | | (k1 k ) k k1 Now we wish to appraise this in a suitable way.

2(k2 + k2) = (k + k2)2 + (k k)2 1 1 1 − (k + k)2 ≥ 1 N2, ≥ from our discussion of adjacent fractions. So

2 2 2k z′ | hk| ≤ N2 √2 k or z′ · ; | hk|≤ N √2 k also z′′ · | hk|≤ N So the inequality becomes 161

1/2 2πih /k 2π/z 2πnz/k2 4 k ∞ (2πν π/12) π( 1 +2/n1)/k2 ′ Ψ z √ ν 2 f (e − ) 1 k( )e 2 1/2 p( )e − e − − ≤ · N = ν 1 1/2X 2π n k Ce | | ≤ N1/2

∞ (2πν π/12) where C is independent of ν, since the series p(ν)e− − is convergent. ν=1 Since the length of the chord of integration < 2P√2 k/(N + 1)m, we have · 3/2 2π n k I∗ < C e | | hk 1 N3/2

17. Lecture 123

Then

iωhk 2πnh/k 2π n 1 ′ e− I∗ C e | | ′ k5/2 hk ≤ 1 kN3/2 0 h

2π n 1 C e | | , ≤ 1 N3/2 0

Hence iωhk 2πinh/k p(n) = e− I + R k5/2 hk N o h

The formula that we had last time looked like this: 163

′ 2πinh/k 5/2 p(n) = iωhke− k− Ihk + RN , o h

zZhk′ and the path of the integration was the arc from zhk′ to zhk′′ in the sense indicated. And now what we do is this. We shall add the missing piece and take the integral over the full circle, how 0 over excluding the origin. Now the path is taken in the negative sense, and we indicate this by writing

2πnz/k2 Ψk(z)e dz.

( ) kZ− This is an improper integral with both ends going to zero. That it exists is

zhk′ clear, for what do we have to compensate for that? we have to subtract 0 ··· R 124 18. Lecture 125

0 and , and we prove that these indeed contribute very little. What is after 164 ··· z′′ Rhk all Ψk(z)? π ( 1 z ) Ψk(z) = √ze 12 3 − k2 0 < Rz 1 and R1/z = 1 on the circle, so that ≤ Ψ (z) √zeπ/12 | k |≤| | 2πnz/k2 2π n and e e | |; | |≤ so that the integrand is bounded. Hence the limit exists. This is indeed very astonishing, for Ψ has an essential singularity at the origin; but on the circle it does not do any harm. Near the origin there are value which are as big as we want, but we can approach the origin in a suitable way. This is the advantage of this contour. The earlier treatment was very complicated.

We can now estimate the integrals. Since zhk′ √2 k/N, the chord can be| a|≤ little longer,· in π fact times the chord at most. 2 So 0

zhk′ 1 2 π k 2π n √z k π/12 √2 e | | · e ≤ · 2 N N ! Z0 2π n 3/2 3/2 Ce | |k N− . ≤ 0 The same estimate holds good for . Hence introducing. ··· z′′ Rhk Page missing page No 165 165

Now everything is under our control. N appeared previously tacitly in zhk′ , 166 because zhk′ depends on the Farey arc. Now N appears in only two places. So p(n) is the limit of the sum which we write symbolically as

∞ 5/2 2πnz/k2 p(n) = i Ak(n)k− Ψk(z)e dz = k 1 Z( ) X K − 18. Lecture 126

(n R 0, integral, and p(n) = 0 for n < 0). So we have an exact inﬁnite series for p(n).

A thing of lesser signiﬁcance is to determine the sum of this series. So we have to speak about the integral. Let us take one more step. Let us get away from 1 the circle. Replace z by w . We 0 do know what this will mean. w will now run on a line parallelto the imaginary axis, from 1 i to 1 + i . So − ∞ ∞

1+i ∞ ∞ 5/2 1/2 π (ω 1/k2ω) 2πn dω = ω 12 k2ω p(n) Ak(n)k− − e − e 2 − = · ω k 1 1 Zi X − ∞ 1+i ∞ 1 ∞ π + π 5/12 5/2 2 (24n 1) = Ak(n)k− ω− e 2 12k ω − dω i = k 1 1 Zi X − ∞ One more step is advisable to get a little closer to the customary notation. 167 πω We then get traditional integrals known in literature. Put = s, 12

π + 12 i 3/2 ∞ 2 1 π ∞ 5/2 5/2 s+ π p(n) = A (n)k− s− e 12k2s (24n 1)ds i 12 k − k=1 π Zi X 12 − ∞ One could look up Watson’s ‘Bessel Functions’ and write down this integral as a Bessel function. But since we need the series anyway we prefer to compute it directly. So we have to investigate an integral of the type

c+i ∞ 1 ν 1 s+ ν L(ν) = s− − e s ds 2πi c Zi − ∞ It does not matter what c > 0 is because it means only a shift to a parallel line, and the integrand goes to zero for large imaginary part of s. For absolute 1 convergence it is enough to have a little more than s− . So take Rν> 0; in our case ν = 3/2. 18. Lecture 127

So let us study the integral

c+i ∞ 1 ν 1 s+ v L (v) = s− − e s ds ν 2πi c Zi − ∞ leaving it to the future what to do with v. The integration is along a 168

line parallel to the imaginary axis. We now bend the path into a loop as in the ﬁgure and push the contour out, so 0 that along the arcs we get negligible contributions.

The contribution from the arc s = R is | | 1 O Rν+1 R · ! s+ ν c Rν/R ν since e s e e , for a ﬁxed v; this is O(R− ) 0 as R , since ν > 0.| So| the ≤ integral along the ordinate becomes a→ ‘loop integra→l’, ∞ starting from along the lower bank of the real axis, looping around the origin and proceeding−∞ along the upper bank towards ; the loop integral is written in a fashion made popular by Watson as −∞

(0+) 1 ν 1 s+ ν s− − e s ds 2πi Z −∞ For better understanding we take a speciﬁc loop. On the lower bank of the 169 negative real axis we proceed only up to ǫ, −

then go round a circle of radius ǫ in the positive sense and proceed thence along the upper bank, the integrand now having acquired a new value-unless ν is an integer. This we take as a standardised loop. We now prove that Lν(V ) is actually diﬀerentiable and that the derivative can be obtained by diﬀerentiating 18. Lecture 128

under the integral sign. For this we take Lν(V + h) Lν(V ) /h and compare V { − } ﬀ it with what we could foresee to be Lν′ ( ) and show that the di erence goes to zero as h 0. → (0+) V + V Lν( h) Lν( ) 1 ν 2 s+ ν − s− − e s ds h − 2πi Z −∞ + (0 ) + ν v h e s ν 1 ν 1 s e s − e s = s− − e ds 2πi h − s Z   −∞    (0+)    h  1 ν 1s+ ν e s 1 1 = s− − e s − ds 2πi h − s Z   −∞     Now   170

h h + h2 + e s 1 1 s s2,s! 1 − = ··· h − s h − s 1 h = h + + s2 2! s3 3! ··· ( · · ) On the path of integration, s ǫ > 0; so | |≥ h 1 e s 1 − C h , h − s ≤ | |

since we are having a quickly converging power series.

(0+) + Lν(ν h) Lν(ν) 1 v 2 s+ v ∴ − s− − e s ds h − 2πi Z −∞ ∞ 1 Rν 1 Rν x ǫ + ǫ = C h 2 + e− e dx 2πǫ + e 0(h) ≤ | |  xv 1 · eν 1   Zǫ    +   Lv(ν h) Lv (ν)  So the limit lim − exists and Lν(ν) is differentiated uniformly in a h 0 h circle of any size.→ Since the differential integral is of the same shape we can differentiate under the integral as often as we please. Lecture 19

The formula for p(n) looked like this: 171

(0+) 3/2 1 π ∞ 5/2 5/2 s+ 1 ( π )2(24n 1) p(n) = A (n)k− s− e s 12k − ds i 12 k k=1 Z X −∞ We discussed the loop integral

(0+) 1 v 1 s+ v L (ν) = s− − e s ds, Rν> 0. v 2πi Z −∞ We can diﬀerentiate under the integral sign and obtain

(0+) 1 v 2 s+ v L′ (ν) = s− − e s ds = L + (ν) v 2πi v 1 Z −∞ This integralis again of the same sort as before; so we can repeat diﬀerenti- ation under the integral sign. Clearly then Lv(ν) is an entire function of ν Lv(ν) has the expansion in a Taylor series: ·

(r) ∞ Lv (0) r Lv(ν) = ν = r! Xr 0 (r) Lv (ν) can be foreseen and is clearly

(0+) 1 v 1 r s+ v s− − − e s ds 2πi Z −∞ 129 19. Lecture 130

So 172 (0+) r ∞ ν 1 v 1 r s L (ν) = s− − − e ds v r! 2πi r=0 Z X −∞ We now utilise a famous formula for the Γ-function - Hankel’s formula, viz, (0+) 1 1 µ s = s− e ds. Γ(µ) 2πi Z −∞ π This is proved by means of the formula Γ(s)Γ(1 s) = and the Euler − sin πs integral. Using the Hankel formula we get L explicitly:

r = ∞ ν Lv(ν) Γ + + = r! (v r 1) Xr 0 What we have obtained is something which we could have guessed earlier. Expanding ev/s as a power series, we have

(0+) r 1 v 1 s ∞ (v/s) L (ν) = s− − e ds v 2πi r! Z r=0 −∞ X (0+) r 1 ∞ v s v 1 r = e s− − − ds, 2πi r! Z r=0 −∞ X and what we have proved therefore is that we can interchange the integration and summation. We have = Lv′ (ν) Lv+1(ν). Having this under control we can put it back into our formula and get a ﬁnal statement about p(n).

3/2 2 π ∞ 5/2 π p(n) = 2π Ak(n)k− L3/2 (24n 1) 12 = 12k − Xk 1 ! This is not yet the classical formula of Hardy and Ramanujan. One trick 173 = one adopts is to replace the index. Remembering that Lv′ (ν) Lv+1(ν), we have π 2 π 2 L / (24n 1) = L′ (24n 1) 3 2 12k − 1/2 12k − ! ! 19. Lecture 131

6k2 d π 2 = L / (24n 1) π2 dn 1 2 12k − ! Let us write the formula for further preparation closer to the Hardy Ra- manujan notation:

1/2 2 π ∞ 1/2 d π p(n) = Ak(n)k− L1/2 (24n 1) 12 = dn 12k − Xk 1 ! 1 Now it turns out that the L-functions for the subscript 2 are elementary functions. We introduce the classical Bessel function

+ ∞ ( )r(z/2)2r v z = v( ) −Γ + + J = r! (v r 1) Xr 0 and the hyperbolic Bessel function (or the ‘Bessel function with imaginary argument’) + ∞ (z/2)2r v z = v( ) Γ + + I = r! (v r 1) Xr 0 How do they belong together? We have 174

z2 z v L = (z) − , v 4 Iv 2 ! z2 z v L = (z) − , v − 4 Jv 2 ! connecting our function with the classical functions. In our case therefore we could write in particular

2 1/2 π π π − L / (24n 1) = / √24n 1 √24n 1 1 2 12k − I1 2 6k − 12k − ! This is always good,but we wouldcomeinto troubleif we have 24n 1 0. It is better to make a case distinction; the above holds for n 1, and for−n ≤ 0, n = m, we have ≥ ≤ − π 2 π 2 L / (24n 1) = L / (24m + 1) 1 2 12k − 1 2 − 12k ! ! π π 1/2 = J √24m + 1 √24m + 1 − 1/2 6k 12k 19. Lecture 132

So we have: n 1. ≥ π 1/2 1/2 √24n 1 π ∞ 1/2 d 6k p(n) = Ak(n)k− I − 12 dn π 1/2 k=1  √24n 1  X  12k −      n = m 0   175 − ≤ π 1/2 J1/2 √24m + 1 π ∞ 1/2 d 6k p(n) = p( m) = Ak( m)k− − − 12 − · dm π 1/2 k=1  √24m + 1  X  12k    We are not yet quite satisﬁed. It is interesting to note that the last expression  is 1 for n = 0 and0 for n < 0. We shall pursue this later. We have now more or less standardised functions. We can even look up tables and compute the Bessel function. However 1/2 and J1/2 are more elementary functions. I

+ ∞ ( )r(z/2)2r 1/2 z = J1/2( ) − 3 = r!Γ(r + ) Xr 0 2 r 2r+ 1 ∞ ( ) (z/2) 2 = − 1 1 1 1 = r! r + r Γ Xr 0 2 − 2 ··· 2 2 1/2 + ∞ rz2r 1 = 2 ( ) − + πz = (2r 1)! ! Xr 0 2 1/2 = sin z. πz ! Similarly if we has abolished ( )r we should have 176 − 2 1/2 I 1 (z) = sinh z 2 πz ! 1/2 I1/2(z) 2 2 sinh z = I1/2(z) = = (z/2)1/2 z √ z ! 2 J (z) z 1/2 = 2 sin (z/2)1/2 √π z We are now at the ﬁnal step in the deduction of our formula: n 1 ≥ 19. Lecture 133

π √ 1 ∞ 1/2 d sinh 6k 24n 1 p(n) = Ak(n)k− − √ dn π √ 3 k=1  6k 24n 1  X  −  π π 2 1  c 1  2 or with √24n 1 = (n ) = n , C = π , 6k − k r3 − 24 k r − 24 r3 c 1 1 ∞ d sinh k n 24 p(n) = A (n)k1/2 − k dn  q  π √2 k=1 1  n 24  X  −   q    n = m 0 − ≤

sin c m + 1 1 ∞ 1 d k 24 p(n) = p( m) = A ( m)k 2 − − k − dm  q  π √2 k=1 + 1  m 24  X    q  This is the final shape of our formula -a convergent series for p(n). 177 The formula can be used for independent computation of p(n). The terms become small. It is of interest to find what one gets if one breaks the series off, say at k = N π5/2 N p(n) = + RN 12 √3 = ··· Xk 1 Let us appraise R A (n) k, because there are only ϕ(k) roots of unity. N ·| k |≤ We want an estimate for L / . For n 1, 3 2 ≥ π2 r 2 π ∞ 6k2 n L3/2 (24n 1) 12k − ≤ r!Γ r + 5 ! r=0 2 X r π2 ∞ 6(N+1)2 n 1 1 3 ≤ = r!Γ r + Xr 0 2 · 2 · 2 π2 r 2r+1 1 ∞ + 2 n 2 > = 6(N 1) · (sincek N in RN ) 3 √π = (2r + 1)!( r + ) Xr 0 2 2π2 r 2 ∞ 3(N+1)2 n + ≤ √π = (2r 1)! Xr 0 19. Lecture 134

2π2 r 2 2 ∞ 3(N+1)2 n

≤ 3 · √π = (2r)! Xr 0 4 π 3n < e N+1 √ 3 3 √π

2 π π 2n ∞ 1 ∴ N+1 √ 3 RN e 3/2 | |≤ 9 √3 = + k kXN 1 2 ∞ π π 2n dk e N+1 √ 3 ≤ 9 √3 k3/2 ZN 2 2π π 1 + √2n3 ∴ RN < e N 1 | | 9 √3 N1/2

This tells us what we have in mind. Make N suitably large. Then one gets 178 something of interest. Put N = [α √n], α constant. Then

1 RN = O(n− 4 )

And this is what Hardy and Ramanujan did. Their work still looksdifferent. They did not have infinite series. They had replaced the hyperbolic sine by the most important part of it, the exponential. The series converges in our case since sinh x x as x 0, so that sinh c n 1 behaves roughly like c . On ∼ → k − 24 k ff 1 q 1/2 3/2 di erentiation we have k2 so that along with k in the numerator we get k− and we have convergence. In the Hardy-Ramanujan paper they had

[ √n] c √n 1 1 d e k − 24 1 = 1/2 + 4 + p(n) Ak(n)k O(n− ) R∗N √ dn  1  2π 2 k=1 n X  24   −   q  sinh was replaced by exp.; so they neglected  179

[ √n] c √n 1 1 1/2 d e− k − 24 R∗ = Ak(n)k √ dn  1  2π 2 k=1 n X  24   −  [ √n] c 1  q c  1 k √n 24  k √n 24 3/2 e− − c e− − R∗ = O k + | |  n 1 · k (n 1 )  k=1  − 24 − 24  X         19. Lecture 135

The exponential is strongly negative if k is small; so it is best for k = 1. Hence

[ √n] [ √n] 1 1/2 1 3/2 R∗ = O k + k | | n   k=1 √n k=1   X X  N       k1/2 = O(N3/2)  = Xk 1 N k3/2 = O(N5/2) = Xk 1 So

1 3/4 1 5/4 R∗ = O n + n | | n √n !! 1 = O n− 4

The constants in the O-term were not known at that time so that numerical 180 computation was difficult. If the series was broken off at some other place the terms might have increased. Hardy and Ramanujan with good instinct broke off at the right place. We shall next resume our function-theoretic discussion and cast a glance at the generating function for p(n) about which we know a good deal more now. Lecture 20

We found a closed expression for p(n); we shall now look back at the generat- 181 ing function and get some interesting results.

1 ∞ f (x) = = p(n)xn, ∞ (1 xn) n=0 n=1 − X Q and we know p(n). p(n) in its simplest form before reduction to the traditional Bessel functions is given by

3/2 2 π ∞ 5/2 π p(n) = 2π Ak(n)k− L3/2 (24n 1) , 12 = 12k − Xk 1 ! 2 π 1 r π 1 ∞ 6k2 (n 24 ) where L3/2 n = − 2 5 6k − 24 = r!Γ + r !! Xr 0 2 We wish ﬁrst to give an appraisal of L and show that the series for p(n) converges absolutely. The series is

π 3/2 ∞ ∞ π2 = π n 5/2 α , f (x) 2 x Ak(n)k− L3/2 2 (n ) 12 = = 6k − Xn 0 Xk 1 ! 1 = where we write 24 α for abbreviation - it will be useful for some other purposes also to have a symbol there instead of a number. We make only a crude estimate.

2 r 2 π π ∞ 6 n L3/2 (n α) 6k2 − ≤ Γ 5 + ! r=0 r! 2 r X 136 20. Lecture 137

2 r π n = ∞ 6 1 1 3 3 = r!Γ + r r 0 2 · 2 · 2 ··· 2 X r 2 2π 2 ∞ 3 πn + + √π = (2r 1)!(3 2r) Xr 0 ∞ (C √n)2r 2 4 , C = π , ≤ (2r)! 3 r=0 r X ∞ (C √n)ρ 4 ≤ = ρ! Xρ 0 = 4eC √n

So f (x) is majorised by 182

∞ ∞ 1 constant x x neC √n 3/2 = | | = k Xn 1 Xk 1 and this is absolutely convergentfor x < 1, indeed uniformly so for x 1 δ, δ > 0, because eC √n = 0(eδn), δ > 0,| | so that we need take xeδ < 1.| |≤ We− can therefore interchange the order of summation: | |

π 3/2 ∞ ∞ π2 = π 5/2 n α f (x) 2 k− Ak(n)x L3/2 2 (n ) 12 = = 6k − Xk 1 Xn 0 ! 3/2 2 π ∞ ∞ h n π 5/2 ′ 2πi = 2π k− ω xe− k L / (n α) 12 hk 3 2 6k2 − k=1 h mod k n=0 X X X ! where the middle sum is a ﬁnite sum. This is all good for x < 1. Now call 183 | | ∞ π2 Φ z = zn k( ) L3/2 2 (n α) = 6k − Xn 0 ! So in a condensed form f (x) appears as

3/2 π ∞ 5/2 2πi h f (x) = 2π k− ′ ω Φ xe− k 12 hk k k=1 h mod k X X We have now a completely new form for our function. It is of great interest to consider Φ (z) for its own sake; it is a power series ( z < 1) and the k | | 20. Lecture 138

coefficients of zn are functions of n α. − ∞ νr L3/2(ν) = 5 = r!Γ + r Xr 0 2 This is an entire function of ν, for the convergence is rapid enough in the whole plane. Looking into the Hadamard theory of entire functions, we could 1 see that the order of this function is . This is indeed plausible, for the de- 2 2r nominator is roughly (2r)! and νr = ( √ν) e √ν; or the function grows (2r)! (2r)! ∼ like e √ν, and this is characteristic of the growth of an entire function or order P P 1 ffi n 2 . The coe cients of z are themselves entire functions in the subscript n. We now quote a theorem of Wigert to the following effect. Suppose that we have a power series Φ(z) = ∞ g(n)zn where g(ν) is in entire function of order 184 n=0 Φ less than 1; then we can sayP something about (z) which has been defined so far for z < 1. This function can be continued analytically beyond the circle of convergence,| | and Φ(z) has only z = 1 as a singularity; it will be an essential singularity in general, but a pole if g(ν) is a rational function. We can extract the proof of Wigert’s theorem from our subsequent arguments; so we do not give a separate proof here.

Φk(z) is a double series : 2 r π (n d) ∞ n ∞ 6k2 Φk(z) = z − , z < 1 5 = = r!Γ + r | | Xn 0 Xr 0 4 This is absolutely convergent; so we can interchange summations and write

2r π ∞ k √6 ∞ r n Φk(z) = (n α) z 5 = r!Γ + r = − Xr 0 2 Xn 0 2r π ∞ k √6 = ϕr(z) 5 = r!Γ + r Xr 0 2 ∞ r n where ϕr(z) is the power series (n α) z . Actually it turnsoutto be a rational n=0 − Φ function. k(z) can be extendedP over the whole plane.

∞ n 1 ϕr(z) = z = . = 1 z Xn 0 − 20. Lecture 139

Diﬀerentiating ϕr(z), 185

= ∞ r n 1 ϕr′ (z) n(n α) z − , = − Xn 0 = ∞ r n zϕr′ (z) n(n α) z , = − Xn 0 ∞ r n αϕr(z) = α(n α) z ; = − Xn 0 so, = ∞ r+1 n = zϕr′(z) αϕr(z) (n α) z ϕr+1(z) − = − Xn 0 This says that we con derive ϕr+1(z) from ϕr(z) by rational processes and diﬀerentiation. This will introduce no new pole; the old pole z = 1 (pole for ϕ (z)) will be enhanced. So ϕr(z) is rational. Let us express the function ◦ = 1 1 + = a little more explicitly in terms of the new variable u z 1 or u 1 z. r+1 r+1 − Introduce ( ) ϕr(z) = ( ) ϕr(1 + u) = ψr(u), say he last equation which was a recursion− formula now− becomes

r+2 1 r 2 r+1 ( ) ψ + (u) = + 1 ( ) u ψ′ (u) α( ) ψ (u) − r 1 u − r − − r ! r+1 1 1 r 1 1 because ψ′ (u) = ( ) ϕ′ 1 + = ( ) ϕ′ 1 + r − r u −u2 − r u u2 ! ! ! = + + ∴ ψr+1(u) u(u 1)ψr′ (u) αψr(u)

This is a simpliﬁed version of our recursion formula. We have a mind to expand about the singularity z = 1. Let us calculate the ψ′ s.

ψ0(u) = u 2 ψ1(u) = u(u + 1) + αu = (1 + α)u + u 2 ψ2(u) = u(u + 1)(2u + 1 + α) + α(1 + α)u + αu = (1 + α)2u + (2α + 3)u3 + 2u3

ψr(u) is a polynomialof degree r+1 without the constant term. The coeﬃcients 186 are a little complicated. If we make a few more trials we get by induction the following: 20. Lecture 140

Theorem. r j r j+1 ψr(u) = ∆ (α + 1) u , = Xj 0 where ∆ j is the jth diﬀerence. By deﬁnition,

∆ f (x) = f (x + 1) f (x), − ∆2 f (x) = ∆∆ f (x) = ∆ f (x + 1) ∆ f (x) − = f (x + 2) 2 f (x + 1) + f (x) − The binomial coeﬃcients appear, and

k k k ℓ k ∆ f (x) = ( ) − f (x + ℓ) = − ℓ Xℓ 0 ! How does the formula for ψr ﬁt? For induction one has to make sure that 187 the start is good.

ψ0(u) = (α + 1)◦u = u 2 2 ψ1(u) = (α + 1)′u′ + ∆(α + 1)′u = (α + 1)u + u 2 2 2 2 2 3 ψ2(u) = (α + 1) u′ + ∆(α + 1) u + ∆ (α + 1) u = (α + 1)2u + (α + 2)2 (α + 1)2 u2 + 2u3 − = (α + 1)2u + (2 α + 3)u2 + 2u3

So the start is good. We assume the formula up to r.

r 2 j r j j r j+1 ψr+1(u) = (u + u)( j + 1)∆ (α + 1) u + α∆ (α + 1) u j=0 X n o r+1 j 1 r j+1 j r j+1 = j∆ − (α + 1) u + ( j + 1 + α)∆ (α + 1) u j=0 X n o (A Seemingly negative diﬀerence need not bother us because it is accompanied by the term j = 0).

r+1 j+1 j 1 r j r = u j∆ − (α + 1) + ( j + 1 + α)∆ (α + 1) j=0 X 20. Lecture 141

+ To show that the last factor is ∆ j(α + 1)r 1, we need a side remark. Intro- 188 duce a theorem corresponding to Leibnitz’s theorem on the diﬀerentiation of a product. We have

∆ f (x)g(x) = f (x + 1)g(x + 1) f (x)g(x) − = f (x + 1)∆g(x) + f (x + 1)g(x) f (x)g(x) − = f (x + 1)∆g(x) + ∆ f (x) g(x) · The general rule is

k k k k ℓ ℓ ∆ f (x)g(x) = ∆ − f (x + ℓ)∆ g(x) = ℓ Xℓ 0 ! This is true for k = 1. We prove it by induction,

∆k+1 f (x)g(x) = ∆(∆k f (x)g(x), and since ∆ is a linear process, this is equal to

k k ∆k ℓ f (x + ℓ + 1)∆ℓ+1g(x) + ∆k+1 ℓ f (x + ℓ)∆ℓg(x) , ℓ − − ℓ=0 X ! n o which becomes, on rearranging summands,

k+1 k+1 ℓ ℓ k k ∆ − f (x + ℓ)∆ g(x) + , = ℓ ℓ 1 Xℓ 0 ( ! − !) k+1 k = k = and the last factor is ℓ , 1 k+1 0 This proves the rule. Applying this to (α + 1)r−, 189 (α + 1)r+1 = (α + 1)(α + 1)r; write f = α + 1, g = (α + 1)r, and observe that f being linear permits only 0th and 1st diﬀerences;

k r k k 1 r k k r ∆ (α + 1) = ∆ − (α + 1) + (α + k + 1)∆ (α + 1) k 1 k − ! ! k 1 r k r = k∆ − (α + 1) + (α + k + 1)∆ (α + 1) r j r j+1 ∴ ψr(u) = ∆ (α + 1) u = Xj 0 20. Lecture 142

We can now rewrite the ϕ′ s:

ϕ (z) = ( )r+1ϕ (n) r − r r 1 = r+1 ∆ j α + r ( ) ( 1) j+1 − = (z 1) Xj 0 −

ϕr has now been deﬁned in the whole plane. Lecture 21

We have rewritten the generating function f (x) as a sum consisting of certain 190 functions which we called Φk(x):

3 π 2 ∞ 5/2 2πi h f (x) = 2π k− ′ ω Φ xe− k 12 hk k k=1 h mod k X X ∞ π2 Φ z = α zn, where k( ) L3/2 2 (n ) = 6k − Xn 0 ! = 1 Φ with α 24 . k(z) could also be written as

2r π ∞ k √6 Φk(z) = ϕr(z′′) 5 = r!Γ + r Xr 0 2 where ϕr(z)is a rationalfunctionas we foundout. We got ϕ explicitly by means of a certain ψ: r 1 ϕ z = r+1 ∆ j α + r r( ) ( ) α( 1) j+1 − = (z 1) Xj 0 − What we need for questions of convergence is an estimate of ϕr; this is not diﬃcult. ∆ f (x) = f (x + 1) f (x) = f ′(ξ ), x < ξ < x + 1, − 1 1 by the mean-value theorem; and because ∆ is a ﬁnite linear process we can interchange it with the operation of applying the mean value theorem and obtain 191

2 ∆ f (x) = ∆(∆ f (x)) = ∆ f ′(ξ ) = f ′(ξ + 1) f ′(ξ ) 1 1 − 1 = f ′′(ξ2), x < ξ1 < ξ2 < ξ1 + 1 < x + 2;

143 21. Lecture 144

3 2 2 ∆ f (x) = ∆(∆ f (x)) = ∆ f ′(ξ), x <ξ< x + 1,

= f ′′′(ξ3), x <ξ<ξ3 < ξ + 2 < x + 3;

and in general ∆k f (x) = f k(ξ), x <ξ< x + k. This was to be expected. Take 1 z δ, 0 <δ< 1 so that z is not too 1 | − | ≥ close to 1. > 1 and 0 <α< 1 δ

r 1 ϕ z + + α + r j r( r(r 1) (r j 1)(1 j) − j+1 | |≤ = − ··· − · δ Xj 0 r (α + 1 + r)r < j+1 = δ Xj 0 (α + 1 + r)r < (r + 1) δr+1 (α + 1 + r)r+1 < δr+1 Originally we know that the formula for f (x) was good for x < 1. From | | this point on we give a new meaning to ϕr(z) for all z , 1. This is a new step. We prove that the series for Φk(z) is convergent not 192 merely for z < 1 but also elsewhere. The sum in Φ (z) is majorised by | | k 2 r π r+1 1 ∞ 6 (α + 1 + r) 5 r δ = r!Γ + r · δ Xr 0 2 This is convergent, for thought the numerator increases with r, we have by Stirling’s formula r r r r r = e + 1 r! r 2 r ∼ √2πr e− √2πr So as far as convergenceis concerned it is no worse than

r eπ2 r 1 ∞ δ α + 1 6 (α + 1 + r) 1 + 5 δ = Γ + r r Xr 1 2 ! which is Cδ, the power series still being rapidly converging because of the ≤ eπ2 + α+1 r factorial in the denominator and 6δ is ﬁxed and 1 r is bounded. So we have absolute convergence and indeed uniformly so for 1 z δ. | − |≥ 21. Lecture 145

We have now a uniformly convergent series outside the point z = 1, and Φk(z) is explained at every point except z = 1 which is an essential singularity. 1 Φ (z) is entire in . From this moment if we put it back into our argument k 1 z − 2πih/k we have f (x) in the whole plane if xe− keep away from 1. And we are sure 193 of that; either x 1 δ or x 1 + δ. Originally x was only inside the unit circle; now it can| | ≤ be outside− also.| | ≥ In both cases f (x) is majorised by

∞ 5 ∞ 3/2 k− 2 , k Cδ = Cδ k− , = · = Xk 1 Xk 1 which is absolutely convergent. Therefore we have now a very peculiar situation. In this notation of Φk we have obtained a function which represents two analytic functions separated by a natural boundary which is full of singularities and cannot be crossed. They are not analytic continuations. The outer function is something new; it is analytic because the series is uniformly convergent in each compact subset. Consider the circle. We state something more explicit which explains the behaviour at each point near the boundary. Since every convergenceis absolute there are no diﬃculties and convergence prevails even if we take each piece separately.

3/2 π ∞ 5 ′ f (x) = 2π k− 2 ωhk − 12 = Xk 1 h Xmod k π2 r ∞ 6k ∞ j r 1 − ∆α(α + 1) 5 2πih/k j+1 = r!Γ + r = · (xe− 1) Xr 0 2 Xj 0 − We can now rearrange at leisure. 194

3/2 π ∞ 5 f (x) = 2π k− 2 ′ ω − 12 hk k=1 h mod k X X 2 r 2πi h ( j+1) π ∞ e− k ∞ j r 6k2 ∆α(α + 1) − (x e2πih/k) j+1 5 = r= j r!Γ + r Xj 0 − X 2 However, if we replaced r∞= j by r∞=0 it would not to any harm because the summation is applied to a polynomialof degree r andthe orderof thediﬀerence is one more than the power. WeP can thereforeP write, taking ∆ outside, 21. Lecture 146

3/2 π ∞ 5 f (x) = 2π k− 2 ′ ω − 12 hk k=1 h mod k X X 2 r 2πi h + π ∞ e k ( j 1) j ∞ r 6k2 ∆α (α + 1) − 2πih/k j+1 5 = (x e ) = r!Γ + r Xj 0 − Xr 0 2 5/2 2πi h ℓ 2 π − ∞ 5 ∞ e k ℓ 1 π = π 2 ′ ω ∆ α + 2 k hk 2πih/k ℓ α− L3/2 2 ( 1) − 12 = = (x e ) −6k Xk 1 h Xmod k Xℓ 1 − ! It is quite clear what has happened. x appears only in the denominator, a root of unity is subtracted and the difference raised to a power 1. Choose 195 B specific h, x, 1; then we have a term . We have a conglomerate of (x e2πih/k)ℓ terms which look like this, a conglomerate− of singularities at each root of unity. So we have a partial fraction decomposition not exactly of the Mittag-Leffler type. Here of course the singularities are not poles, and they are everywhere dense on the unit circle. Each series ∞ represents one specific point e2πih/k. ℓ=1 Let us return to our previous statement.P f (x) is regular and analytic outside the unit circle. What form has it there? Inside it is ∞ (1 xm). We shall expand m=1 − f (x) about the point at infinity. We want the ϕ′ s explicitly.Q 1 ϕ (z) = 0 1 z − ϕ + (z) = zϕ′ (z) αϕ (z) r 1 r − r z 1 z 1 ∞ ϕ z = − = − = z m 0( ) 1 1 − z− 1 −1 z− − = − − Xm 1 ∞ m ∞ m ∞ m ϕ1(z) = mz− + α z− = (m + α)z− = = = Xm 1 Xm 1 Xm 1 The following thing will clearly prevail

r+1 ∞ r m ϕr(z) = ( ) (m + α) z− − = Xm 1 This speaks for itself.

r ∞ m r ∞ r m ϕr+1(z) = ( ) m(m + α)z− + ( ) α (m + α) z− − = − = Xm 1 Xm 1 21. Lecture 147

So the general formula is justiﬁed by induction. 196

π2 r ∞ 6k2 ∞ r m Φk(z) = − (m + α) z− 5 − = r!Γ + r = Xr 0 2 Xm 1 for all z > 1. Exchanging summations, | | 2 r π ( m α) ∞ m ∞ 6k2 Φk(z) = z− − − 5 − = = r!Γ + r Xm 1 Xr 0 2 ∞ π2 = z m α − L3/2 2 ( m ) − = 6k − − Xm 1 ! Put this back into f (x); we get for x > 1, | | 3/2 2 π ∞ 5 ∞ h m π ′ 1 2πi f (x) = 2π k− 2 ω x− e k L / ( m α) , − 12 hk 3 2 6k2 − − k=1 h mod k m=1 X X X ! ′ 2πih/k and since Ak(n) = ωhke− , h Xmod k π 3/2 ∞ ∞ π2 = π 5/2 m α f (x) 2 k− Ak( m)x− L3/2 2 ( m ) − 12 = = − 6k − − Xk 1 Xm 1 ! Again interchanging summations, 197

π 3/2 ∞ ∞ π2 = π m 5/2 α f (x) 2 x− Ak( m)k− L3/2 2 ( m ) − 12 = = − 6k − − Xm 1 Xk 1 ! The inner sum we recognize immediately; it is exactly what we had for p(n); so 3/2 π ∞ m f (x) = 2π p( m)x− − 12 = − Xm 1 And here is a surprise which could not be foreseen! By its very meaning p( m) = 0. So − f (x) 0 ≡ outside the unit circle. This was first conjectured by myself and proved by H.Petersson by a completely different method. Such expressions occur in the theory of modular forms. Petersson got the outside function first and then the inner one, contrary to what we did. 21. Lecture 148

The function is represented by a series inside the circle, and it is zero outside, with the circle being a natural boundary. There exist simpler examples of this type of behaviour. Consider the partial sums:

x 1 1 + = 1 x 1 x − − x x2 1 x2 1 1 + + = + = 1 x (1 x)(1 x2) 1 x (1 x)(1 x2) (1 x)(1 x2) − − − − − − − − x x2 x3 1 + + + + to n + 1 terms 1 x (1 x)(1 x2) (1 x)(1 x2)(1 x3) ··· − − − − − − = 1 (1 x)(1 x2) (1 xn) − − ··· − 1 For x < 1, the partial sum converges to . For x > 1 also 198 | | | | ∞ (1 xm) m=1 − it has a limit; the powers of x far outpace 1 andQ so the denominator tends to inﬁnity and the limit is zero. The Euler series here is something just like our complicated function. Actually the two are the same. For suppose we take the 1 partial sum and break it into partial fractions. We (1 x)(1 x2) (1 xn) get the roots of− unity in− the··· denominator,− so that we have a decomposition

Bh,k,l,n ℓ 2πi h x e k X − k n and ℓ not too high. For a higher n we get a ﬁner expression into partial fractions.≤ Let us face one of these, keeping h, k, ℓ ﬁxed:

Bh,k,l,n ℓ 2πi h x e k − Let n . Then I have the opinion that →∞ 3/2 2 π 5 2πi h ℓ ℓ 1 π B , , , 2π ω k− 2 e k ∆ − L (α + 1) h k l n →− 12 hk α 3k −6k2 ! The B′s all appear from algebraic relations and so are algebraic numbers - in sufficiently high cyclotomic fields. And this is equal to something which looked highly transcendental! though we cannot vouch for this. The verifi- cation is difficult even in simple cases - and no finite number of experiments would prove the result. 21. Lecture 149

B0,1,1,n is itself very complicated. Let us evaluate the principal formula for 199 x 1 f (x) and− pick out the terms corresponding to h = 0, k = l, ℓ = 1. √ 6 12 3 1 = L3/2 is just the sine function and terns out to be . Since 1 x −25 − 75π − 1 x 1 , 1 is the ﬁrst approximation to B0,1,1,n. If we take the partial fraction −decomposition− − for 1 1 , = ·· + ·· + ·· , (1 x)(1 x2) (1 x)(1 x2) (x 1)2 (x 1) (1 + x) − − − − − − the numerator of the second term would give the second approximation. If indeed these successive approximations converge to B0,1,1,n we could get a whole new approach to the theory of partitions. We could start with the Euler series and go to the partition function. We are now more prepared to go into the structure of ωhk. We shall study next time the arithmetical sum Ak(n) and the discovery of A.Selberg. We shall then go back again to the η-function. Lecture 22

We shall speak about the important sum Ak(n) which appeared in the formula 200 for p(n), deﬁned as ′ 2πinh/k Ak(n) = ωhke− . h Xmod k we need the explanation of the ωhk; they appeared as factors in a transformation formula in the following way:

+ + 2πi h iz π 1 z 2πi h′ i/z f e k = ωhk √ze 12k z − f e k , hh′ + 1 0 (mod k) ≡ Here, as we know,

= 1 f (x) m ∞= (1 x ) m 1 − ∞ and as η(τ) = eQπiτ/12 (1 e2πimτ), − m=1 2πiτ πiτ/Y12 1 f (e ) = e (η(τ))−

We know how η(τ) because. ωhk is something belonging to the behaviour of the modular form η(τ). What is ωhk explicitly? We had a formula

aτ + b cτ + d η = ǫ η(τ), c > 0, cτ + d i ! r and 201 epsilonn is just the question. Our procedure will be to study ǫ and η and then go back to f where ωhk appeared. The trick in the discussion will be that we

150 22. Lecture 151

shall not use the product formula for η(τ), but the inﬁnite series from the pentagonal numbers theorem. This was carried out at my suggestion by W.Fischer (Paciﬁc Journal of Mathematics; vol. 1). However we shall not copy him. We shall make it shorter and dismiss for our purpose all the long and complicated discussions of Gaussian sums

k 2 G(h, k) = e2πiν h/k = Xv 1 which are of great interest arithmetically, having to do with law of reciprocity to which we shall return later. We are able to infer that a formula of the sort quotedfor η should exist from V the discussion of 1′(0/τ). We had the formula (see hechire 14) aτ + b V 0 = 1 cτ + d ··· . ! where the right side contains a doubtfulroot of unity, which we could discuss in some special cases, and by iteration in all cases. We shall use as furtherbasis of our argument that such a formula has been established with the proviso ǫ = 1. We then make a statement about ǫ and use it directly. | | After all this long talk let us go to work. We had τ′ = (h′ + i/z/k), τ = (h + iz)/k. The question is how is τ′ produced from τ? It was obtained by 202 means of the substitution

hh′+1 a b h′ = − k c d  k h     −      We can therefore get what we are after if we specify the formula by these particular values. h + iz h + iz η ′ = ǫ √zη k k ! ! with the principal value for √z. We wish to determine ǫ deﬁned by this. We shall expand both sides and compare the results. For expansion we do not use the inﬁnite product but the pentagonal numbers formula.

πiτ/12 ∞ λ 2πiτλ(3λ 1)/2 η(τ) = e ( ) e − λ= − X−∞ ∞ λ πiτ (1+36λ2 12λ) = ( ) e 12 − λ= − X−∞ 22. Lecture 152

∞ λ 3πiτ(λ 1/6)2 = ( ) e − λ= − X−∞ Most determinations of η(τ) make use of the inﬁnite product formula; the inﬁnite series is simpler here

+ + h iz ∞ λ 3πi h iz (λ 1/6)2 η = ( ) e k − k λ= − ! X−∞ In order to get the root of unity a little more clearly exhibited, we replace λ 203 mod 2k. λ = 2kq + j, j = 0, 1,..., 2k 1 and q runs from to . So − −∞ ∞ + 2k 1 h iz ∞ − j 3πi h (2kq+ j 1 )2 3π z (2kq+ j 1 )2 η = ( ) e k − 6 e− k − 6 k − q= j=0 ! X−∞ X = 1 h The product term in the exponent 4kq( j 6 ).3πi k = 2πihq(6− j 1) = an integral− multiple of 2πi

(Thisis the reason whywe used mod 2k).

2k 1 + − h 1 2 ∞ j 1/6 2 h iz j 3πi ( j ) 12πzk(q+ − ) η = ( ) e k − 6 e− 2k . k − j=0 q= ! X X−∞ We did this purposely in order to make it comparable to what we did in the theory of V -functions. For Rt > 0, we have

∞ πt(q+α)2 1 ∞ π m2 2πimα e− = e− t e q= √t m= X−∞ X−∞ This is a consequence of a V -formula we had:

2 1 1 eπiτν V (ντ/τ) = V ν/ 3 τ 3 − τ r ! If we write this explicitly, 204

∞ πiτn2 2πinν V3(ν/τ) = e e , n= X−∞ 22. Lecture 153

and put iτ = t, −

πtν2 ∞ πtn2 2πntν 1 ∞ πn2/t 2πinν e− e− e− = e− e , n= √t n= X−∞ X−∞ ∞ πt(ν+n)2 1 ∞ π n2 2πinν or e− = e− t e , n= √t n= X−∞ X−∞ which is the formula quoted. We now apply this deep theorem and get some- j 1/6 thing completely new. Putting t = 12zk and α = − , k

2k 1 + 2 2 h iz − j 2πi h ( j 1 ) 1 ∞ πm πim ( j 1 ) η = ( ) e k − 6 e− 12zk e k − 6 k − j=0 √12kz m= ! X X−∞ We rewrite this, emphasizing the variable and exchanging the orders of summation. Then

2k 1 2 + z ∞ πm2 − + 2h 1 + m h i 1 πi j k ( j 6 ) 12k (6 j 1) η = e− 12kz e − − k 2 √3kz m= j=0 ! X−∞ X 205 Let us use an abbreviation.

h + iz 1 ∞ πm2 η = e− 12kz T , k (m) √2kz m= ! X−∞ 2K 1 1 − πi( j+ 2h ( j 1 )2+ m (6 j 1)) where T(m) = e k − 6 12k − . 2 = Xj 0 h + iz 1 ∞ πm2 η = T(0) + e− 12kz (T(m) + T( m)) k √3kz − !  m=1   X  1   This is a function in z . Also 

+ 1 h iz ǫ− ∞ λ π (6λ 1)2 πih′ (6λ 1)2 η = ( ) e− 12kz − e 12k − k z √ λ= − ! X−∞ h+iz ﬀ Now η k hasbeen obtainedin two di erent ways. We have in both cases π/(12zk) = a power series in e− x, both for x < 1. But an analytic function has only one power series; so they are identical.| | This teaches us something. The 22. Lecture 154

second teaches us that by no means do all sequences appear in the exponent. 206 Only m2 = (6λ 1)2 can occur. There is no constant term in the second expression. So m has− the form 6λ 1 = 6λ 1, λ > 0. Make the comparison; the coeﬃcients are identical.| They− are| almost± always zero. In particular T(0) = 0. T(m) for m other than 1 (mod 6) also vanish. So we have the following iden- tiﬁcation. ±

1 1 λ πih′ 2 (T(6λ 1) + T( 6λ + 1)) = ǫ− ( ) e 12k (6λ 1) √3k − − − − Realise that we have acknowledged here that a transformation formula exists. The root of unity ǫ is independent of λ. This we can assume but W. Ruscher does not. Take in particular λ = 0. Then we have for m = 1, ±

1 1 πih′ (T( ) + T(1)) = ǫ− e 12k √3k − This is proved by Fischer by using Gaussian sums. Therefore

πih′ 2k 1 2k 1 12k − 3h i 2 6 j 1 − 3h 1 2 6 j 1 1 e− πi( j+ ( j ) ) − πi j+ ( j ) + − ǫ− = e k − 6 − 6k + e k − 6 6k √3k    j=0 j=0  X X    Now j mattersonly mod 2k. We can beautify things slightly: 

πih′ + πih 1 e− 12k 12k πi πi (3h j2+ j(k h 1)) πi πi (3h j2+ j(k h 1)) ǫ− = e 6k e k − − + e− 6k e k − − 2 √3k    j mod 2k j mod 2k   X X    The sum appears complicated but will collapse nicely; however compli- cated it should be a root of unity. In Ak(n) the sums are summed over h and for that purpose we shall not need to compute the sums explicitly. Lecture 23

Last time we obtained the formula 207

πi(h h ) πi πi 2 1 1 − ′ 3h j (k h+1) ǫ− = e 12k e− 6k e k j − 2 √3k j Xmod 2k πi(h h ) πi πi 2 1 − ′ (3h j + j(k h 1)) + e 12k e 6k e k − − 2 √3k j Xmod 2k ωh,k was deﬁned by means of the equation

+ + 2πi h iz π 1 z 2πi h′ i/z f e k = ωhk √ze 12k z − f e k ωhk came from the ǫ in the transformation formula aτ + b cτ + d η = ǫ η(τ) cτ + d i ! r In particular, h + 1/z h + iz η ′ = ǫ √zη , k k ! ! 2πiτ πiτ/12 1 f e = e (η(τ))− Substituting in the previous formula,

1 1 + + + + πi h iz h iz − π ( 1 z) πi h′ iz h′ i/z − e 12 η = ω √ze 12k 3 − e 12 k η k k hk k ( !) ( !) + + h′ i/z πi (h h) h iz i.e., η = ω √ze 12k ′− η k hk k ! ! 155 23. Lecture 156

πi (h′ h) ∴ ǫ = ωhke 12k − π (h′ h) or ωhk = ǫe− 12k −

208 1 In the ﬁrst formula we have obtained an expression for ǫ− . However, we 1 could make a detour and act ǫ directly instead of ǫ− . Even otherwise this could 1 be ﬁxed up, for after all it is a root of unity. We have ǫǫ¯ = 1 or ǫ = ǫ¯− . So consistently changing the sign in the exponents, we have

πi 1 πi πi 2 1 (h h′) (3h j + j(k h+1)) ωhk = ǫ¯− e 12k − = e 6k e− k − 2 √3k j Xmod 2k 1 πi πi (3h j2+ j(k h 1)) + e− 6k e− k − − 2 √3k j Xmod 2k

We now have the ωhk that we need. But the ωhk are only of passing interest; we put them back into Ak(n);

′ 2πinh/k Ak(n) = ωhke− h Xmod k This formula has one unpleasant feature, viz. (h, k) = 1. But this would not do any harm. We can use a lemma from an unpublished paper by Whiteman which status that if (h, k) = d > 1, then

πi (3h j2+ j(k h 1)) e− k − ± = 0 j Xmod 2k

For proving Whiteman status put h = dh∗, k = dk∗ and j = 2k∗l + r, 209 0 1 d 1, 0 r 2k 1. Then ≤ ≤ − ≤ ≤ ∗ − d 1 2k∗ 1 πi 2 − − πi 2 (3h j + j(k h 1)) (3dh∗(2k∗1+r) +(2k∗ℓ+r)(dk∗ dh∗ 1))) e− k − ± = e− dk∗ − ± = = j Xmod 2k Xℓ 0 Xr 0 2k∗ 1 d 1 − πi (3hr2+r(k h 1)) − 2πiℓ/d = e− k − ± e∓ , r=0 = X Xℓ 0 and the inner sum = 0 because it is a full sum of roots of unity and d , 1. This simpliﬁes the matter considerably. We can now write

1 πi πi (3h j2+ j(k h+1)) 2πin h Ak(n) = e 6k e− k − e− k 2 √3k h Xmod k j Xmod 2k 23. Lecture 157

1 πi πi (3h j2+ j(k h 1)) 2πin h + e− 6k e− k − − e− k 2 √3k h Xmod k j Xmod 2k Rearranging, this gives 210

πi πi 2πi j(3 j 1) 1 (k+1) j (n+ − )h Ak(n) = e 6k e− k e− k k 2 √3k j Xmod 2k h Xmod k πi πi 2πi j(3 j 1) 1 (k 1) j (n+ − )h + e− 6k e− k − e− k 2 2 √3k j Xmod 2k h Xmod k The inner sum is equal to the sum of the kth roots of unity, which is 0 or k, k if all the summands are separately one, i.e., if j(3 j 1) n + − 0 (mod k) 2 ≡ Hence

1 k πi j πij 1 k πi j πij A (n) = e 6k ( ) e− k + e− 6k ( ) e k k 2 3 − 2 3 − r j mod 2k r j mod 2k j(3 j 1) X j(3 j 1) X − n (mod k) − n (mod k) 2 ≡− 2 ≡−

In the summation here we ﬁrst take all j′s modulo2k (this is the ﬁrst sieving out), and then retain only those j which satisfy the second condition modulo k (this is the second sieving out). Combining the terms,

1 k j πi (6 j 1) πi (6 j 1) A (n) = ( ) e− 6k − + e 6k − k 2 3 − r j mod 2k j(3 j 1) X − n (mod k) n o 2 ≡− k π(6 j 1) = ( ) j cos − 3 − 6k r j mod 2k j(3 j 1) X − n (mod k) 2 ≡− 211 This formulais due to A.Selberg. It is remarkablehow simple it is. We shall change it a little, so that it could be easily computed. We shall show that the Ak(n) have a certain multiplicative property, so that they can be broken up into prime parts which can be computed separately. Let us rewrite the summation condition in the following way.

12 j(3 j 1) 24n (mod 24k) − ≡− 23. Lecture 158

i.e., 36 j2 12 j + 1 1 24n (mod 24k) − ≡ − i.e., (6 j 1)2 ν (mod 24k) − ≡ where we have written ν = 1 24n. In the formula − 1 k j πi (6 j 1) πi (6 j 1) A (n) = ( ) e− 6k − + e 6k − k 2 3 − r j mod 2k j(3 j 1) X − n (mod k) n o 2 ≡− replace j by 2k j in the popint term (where j runs through a full system of − residues, so does 2k j). Further, observe that we have now 212 − (12k 6 j 1)2 (mod 24k) − − ≡ i.e., (6 j + 1)2 (mod 24k) ≡ Then

1 k j πi (6 j+1) j πi (6 j 1) Ak(n) =  ( ) e− 6k + ( ) e 6k −  2 3  − −  r  j mod 2k j mod 2k  (6 j 1)X2 ν (mod 24k) (6 j 1)X2 ν (mod 24k)   − ≡ − ≡    In both terms the range of summation is j mod 2k and there is the further condition which restricts j. So

1 k j πi (6 j 1) A (n) = ( ) e− 6k ± k 2 3 − r j mod 2k (6 j 1)2 νX(mod 24k) ± ≡ Write 6 j 1 = ℓ. 6 j 1 thus modulo 24k. j = ℓ+1 , so it is the integer ± ± 6 ℓ nearest to ℓ since (ℓ, 6) = 1. So write j = where x denotes the integer 6 6 { } nearest to x. Then ( )

1 k ℓ πiℓ A (n) = ( ) 6 e 6k k 2 3 − { } r ℓ mod 2k (ℓ,6)=1,ℓ2 Xν (mod 24k) ≡ And one ﬁnal touch. The ranges for ℓ in the two conditions are modulo 12k 213 and modulo 24k. Make these ranges the same. Then

1 k ℓ πiℓ A (n) = ( ) 6 e 6k k 4 3 − { } r ℓ mod 24k ℓ2 ν X(mod 24k) ≡ 23. Lecture 159

We prefer the formula in this form which is much handler. We shall utilise this to get the multiplicative property of Ak(n). Lecture 24

We derived Selberg’s formula, and it looked in our transformation like this: 214

1 k ℓ πiℓ A (n) = ( ) 6 e 6k , k 4 3 − { } r l2 γ (mod 24k) ≡ X

where ν = 1 24n, or ν 1 (mod 24). We write this Bk(ν); this is deﬁned for ν 1 (mod− 24), and we≡ had tacitly (ℓ, 6) = 1. We make an important remark ≡ ℓ about the symbol ( ) 6 . This repeats itself for ℓmodulo12. The values are − { } ℓ = 1 3 7 11 ℓ 6 = ( ){ } 1 1 1 1 − ℓ − − But ( ) 6 can be expressed in terms of the Legendre symbol: − { } ℓ ℓ 1 ( ) 6 = − − { } 3 ℓ ! ! ℓ 1 = 1 = −2 when (ℓ, 6) 1. We can test this, noticing that −ℓ ( 1) . Since 1, 7 are quadratic residues and 5, 11 quadratic non-residues mod−ulo 3, we have for ℓ ℓ = 1, 5, 7, 11, ( ){ 6 } = 1, 1, 1, 1 respectively; this agrees with the previous − − − ℓ list. It is sometimes simpler to write ( ){ 6 } in this way, though it is an after- thought. It shows the periodicity. − Let us repeat the formula: 215

1 k ℓ 1 πiℓ B (ν) = − e 6k k 4 3 3 ℓ r ℓ2 ν (mod 24k) ! ! ≡ X This depends upon how k behaves with respect to 24. It has to be done separately for 2, 3, 4, 6. For this introduce d = (24, k3). We have

160 24. Lecture 161

d = 1if (k, 24) = 1, 3if 3 4, k odd, | 8 if k is even and 3 ∤ k 24if 6 k. | Let us introduce the complementary divisor e, de = 24. So e = 24, 8, 3 or 1. (d, e) = 1. Also (c, k) = 1. All this is a preparationfor our purpose. The congruence ℓ2 ν (mod 24k) can be re-written separately as two congruences: ℓ2 ν (mod≡ dk), ℓ2 ν (mod e). ≡ ≡ The latter is always fulfilled if (ℓ, 6) = 1. Now break the condition into two subcases. Let r be a solution of the congruence (er)2 ν (mod dk); ≡ then we can write ℓ = er + dk j, where j runs modulo e and moreover ( j, e) = 1. To different pairs modulo dk and e respectively belong different ℓ modulo 24k. Bk(ν) can then be written as + 1 k er dk j 1 πi (er+dk j) B (ν) = − e 6k k 4 3 3 er + dk j r (er)2 ν (mod dk) j mod e ! ! ≡ X ( jX,e)=1

Separating the summations, this gives 216

1 k πiℓk B (ν) = e 6k S (r), k 4 3 k r (er)2 ν (mod dk) ≡ X where er + dh j 1 πidj S (r) = ′ − e 6k k 3 er + dk j j Xmod e ! ! We compute this now in the four diﬀerent cases implied in the possibilities d = 1, 3, 8, 24. Case 1. d = 1,e = 24

k j 1 πij S (r) = ′ − e 6 k 3 k j j Xmod 24 ! ! k 1 j j 1 πij = − ′ ( ) −2 e 6 3 k 3 − ! ! j Xmod 24 24. Lecture 162

There are eight summands, but eﬀectively only four, because they can be folded together.

k 1 j π 1 πi j S (r) = 2 − ′ ( ) −2 e k 3 k 3 − ! ! j Xmod 12 h 1 πi 5πi 7πi 11πi = 2 − e 6 e 6 e 6 + e 6 3 k − − ! ! n o ℓ (We replaced the nice symbol ( ){ 6 } by the Legendre symbol because we did not know a factorisation law for− the former. So we make use of one special character that we know). k 1 π 5π S (r) = 4 − cos cos k 3 k 6 − 6 ! ! ! k 1 = 4 − √3 3 k ! ! k 3 k 1 1 1 and since = ( ) −2 = − , this gives gives 217 3 k − · k 3 S (r) = 4 √3 k k ! Case 2. d = 3,e = 8.

8r 1 πij S (r) = ′ − e 2 k 3 3k j j Xmod 8 ! ! r 1 1 πij = − − ′ − e 2 3 3k j ! j Xmod 8 ! r 1 1 πij = 2 − ′ − e 2 3 k j ! j Xmod 4 ! r 1 = 2 − (i + i) 3 k ! r 1 = 4i − . 3 k ! Case 3. d = 8,e = 3.

8k j 1 4πij S (r) = ′ − e 3 k 3 3r j Xmod 3 ! ! 24. Lecture 163

k 1 j 4πi = − ′ e 3 3 r 3 ! ! j Xmod 3 k 1 4πi 8πi = − e 3 e 3 3 r − ! ! k 1 2π = 2i − sin − 3 r 3 ! ! 1 k 1 = √3 − i 3 r ! ! 218 Case 4. d = 24,e = 1.

k 1 3 S (r) = − = k 3 r r ! ! !

Now utilise these; we get a handier deﬁnition for Ak(n). Case 1. √k 3 4πir B (ν) = e k k k ! (24r)2 ν (mod k) ≡X Case 2. k 1 r 4πir B (ν) = i − e 3k k 3 k 3 r ! (8k)2 ν (mod 3k) ≡ X The i should not bother us because r and r are solutions together, so they combine to give a real number. −

k 1 r 4πr B (ν) = − sin k − 3 k 3 3k r ! (8r)2 v (mod 3k) ≡ X Case 3.

1 k 1 πir B (ν) = √k − e 3k k 4i 3 r ! (3k)2 ν (mod 8k) ! ≡ X 1 k 1 πr = √k − sin 4 3 r 2k ! (3r)2 ν (mod 8k) ! ≡ X 219 24. Lecture 164

Case 4. 1 k 3 πir B (ν) = e 6k k 4 3 r r r2 ν (mod 24k) ≡ X ! This is the same as the old deﬁnition. This makes it possible to compute Ak(n). We break k into prime factors and because of the multiplicative property which we shall prove, have to face only the task of computing for prime powers. Lecture 25

We wish to utilise the formula for Bk(ν) that we had: 220

1 k ℓ 1 πiℓ A (n) = B (ν) = − e 6k , k k 4 3 3 ℓ r ℓ2 ν (mod 24k) ! ! ≡ X with ν = 1 24n (and so 1 modulo 24). Some cases were considerably simpler. Writing− d = (24, k3≡), de = 24, we have four cases: d = 1, 3, 8, 24. d = 1 3 B (ν) = √k e4πir/k k k ! (24r)2 ν (mod k) ≡X d = 3 1 k r B (ν) = 2i − e4πir/3k k k 3 3 ! r (8r)2 ν (mod 3k) ≡ X d = 8 1 k 1 B (ν) = √k − eπir/2k k 4i 3 r (8r)2 ν (mod 8k) ! ≡ X ! d = 24 There is nothing new; we get the old formula back. We wish ﬁrst to anticipate what we shall use later and get An(n) for prime powers which will be the ultimate elements. Again we have to discuss several cases. First take k = pλ, p a prime exceeding 3. Then, by case 1 above (since 221 (24, k3) = 1), λ 3 λ B (ν) = pλ/2 e4πir/p k p ! (24r)2 ν (mod pλ) ≡X 165 25. Lecture 166

Look into the condition of summation. It is quite clear that this implies (24r)2 ν (mod p) i.e., ν is a quadratic residue modulo p. Hence ≡ v B λ(ν) = 0 if = 1. (1) p p − ! On the other hand, if x2 ν (mod p) is solvable, then x2 ν (mod pλ) is also solvable (we take for granted≡ the structure of the cyclic residue≡ group). x2 ν (mod pλ) has two solutions, and now we want only x = 24r (mod pλ). Let≡r be a solution, r is the other solution: (24r)2 (mod pλ). Then − ≡ λ 3 λ/2 4πir/pλ 4πir/pλ B (ν) = p e + e− k p ! λ n o 3 4πr = 2 pλ/2 cos (2) p pλ ! This is roughly of the order of pλ Next, suppose that p/ν. This is a special case of pλ/ν. Then (24r)2 0 p (mod pλ), and the solutions are ≡

λ+1 r = p[ 2 ] j, · λ [ λ+1 ] j = 0, 1, 2,..., p − 2 1. − = λ+1 = when λ 1, 2 λ and we have only one summand. Hence 222 h i 3 B (ν) = p1/2 (3) k p ! Now let λ> 1. Then

+ λ λ 1 λ p −[ 2 ] 3 λ 4πi j/p[ λ+1 ] Bk(ν) = p 2 e 2 p = ! Xj 1 This again involves two cases, λ even and λ odd. If λ is even, λ = 2µ and the sum becomes pµ µ e4πi j/p = Xj 1 and this is 0, being a full sum of roots of unity. Hence in this case

Bk(ν) = 0 (4) 25. Lecture 167

Now let λ be odd: λ = 2µ + 1.

r = pµ+1 j, j = 0, 1,..., pµ 1. · − Then the sum becomes pµ µ e4πi j/p = Xj 1 which is again zero; hence Bk(ν) = 0 (5)

Now suppose that pµ ν, µ < λ and pλ ∤ ν. r2 ν (mod pλ)ν = pµν, p + ν, 223 1 µ λ | µ 2 ≡ µ λ or ν p ν (mod p ) ν = p ν1, p ∤ ν1; or ν p ν1 (mod p ). If is odd, µ < λ,≡ then pµ ν; and again ≡ | Bk(ν) = 0 (6) There remain the case in which µ is even, µ = 2ρ. Then r2 p2ρν, λ ρ 2ℓ 2 2ρ λ 2 ≡ λ 2ρ (mod p ). Writing r = p j, p j p ν1 (mod p ), or j ν1 (mod p − ) ν1 = ≡ ≡ If p 1, then again − B (ν) = 0 (7) k ν1 = 2 λ 2ρ However p 1 implies j ν1 (mod p − ) has two solutions, j and j. Then ≡ − ρ λ 2ρ λ r p j + ℓp − (mod p ) ≡ ρ λ ρ λ or τ r p j + ℓp − (mod p ) ≡ where ℓ = 0, 1,..., pρ 1. − Then the sum becomes

pρ 1 pρ 1 − 4πi 4πi j − 4πi pλ ρ λ ρ pλ ρ pρ ℓ e ( p j + ℓp − ) = e± − e = ± = Xℓ 0 Xℓ 0 = 0

Again Bk(ν) = 0 (8)

We now take up the case p = 3. This corresponds to p = 3. If k = pλ = 3λ, 224

λ λ 1 r 4πir/3λ+1 B λ(ν) = i( ) 3 −2 e 3 − 3 (8r)2 (mod 3λ+1) ≡ X 25. Lecture 168

ν 1 (mod 24)or ν 1 (mod 3). So ν = 1. There are two solutions, r and ≡ ≡ 3 2 λ+1 r r r for the congruence (8r) ν (mod 3 ). Since − = , − ≡ 3 − 3

r λ 1 4πir 4πir λ − λ+ λ+ B λ(ν) = i( ) 3 2 e 3 1 e− 3 1 3 − 3 − λ+1 r λ 1 4πr = 2( ) 3 −2 sin (9) − 3 3λ+1 Finally, we take p = 2; then d is 8. Let k = 2λ. Then

1 1 λ+1 B λ(ν) = ( )λ2λ/2 − e4πir/2 2 4i − r (3r)2 ν (mod 2λ+3) ! ≡ X ν 1 (mod 8) implies that (3r2) ν (mod 8) has four solutions, and these solutions≡ are inherited by the higher≡ powers of the modulus. The solutions are r 1, 3, 5, 7 (mod 8). In general the congruence x2 ν (mod 2µ), µ 3 has four≡ solutions ≡ ≥ µ 1 r + h2 − , h = 0, 1 ± Then 225

1 λ λ/2 4πir/2λ+1 4πir/2λ+1 4πir/2λ+1 4πir/2λ+1 1 B λ (ν) = ( ) 2 e e− + e e− − 2 4i − − − r n o ! 1 r 1 and since − = ( ) −2 , r − λ λ/2 1 4πr B λ (ν) = ( ) e − sin (10) 2 − r 2λ+! ! We have thus computed the fundamental cases explicitly. Lecture 26

We had the formula for Bk(ν): 226

1 k ℓ 1 B (ν) = − eπiℓ/6k, k 4 3 3 3 r ℓ2 ν (mod 24k) ! ! ≡ X with ν 1 (mod 24). Writing d = (24, k3), we had the following cases: ≡ 1) d = 1 3 B (ν) = √k e4πir/k k k ! (24π)2 ν (mod k) ≡X 2) d = 3 1 k 1 B (ν) = i − − eπir/2k k k 3 r ! r (24r)2 ν (mod 3k) ! ≡X 3) d = 8 1 k 1 B (ν) = √k − eπir/2k k 4i 3 r ! (3r)2 ν (mod 8k) ! ≡ X 4) d = 24. We do not get anything new. = = = Assume k k1k2,(k1, k2) 1. We desire to write Bk(ν1). Bk2 (ν2) Bk(ν), with a suitable ν to be found out from ν1 and ν2. It cannot be foreseen. It is a multiplication of a peculiar sort. Two cases arise.

(i) At least one of k1, k2 is prime to 24 and therefore to 6, say (k1, 6) = 1. 227

(ii) None is prime to 6. But since (k1, k2) = 1, 2/k1, 3/k1. Under the circumstances prevailing these are the two mutually exclusive cases.

169 26. Lecture 170

Case 1. Utilise d = 1.

3 1 k2 = 4πir/k1 Bk1 (ν1) Bk2 (ν2) k1 e · k1 · 4 3 2 ! r (24r) ν1 (mod k1) p ≡ X ℓ 1 − eπiℓ/6k2 · 2 3 ℓ ℓ ν2 (mod 24k2) ! ! ≡ X πi 1 3 k1k2 (24k2r+k1l) l 1 = e 6k1k2 − 4 k1 3 2 3 l ! r (24r) ν1 (mod k1) ! ! ℓ2 νXX≡ (mod 24k ) ≡ 2 2 k1 and 24k2 are coprime moduli. If r runs modulo k1 and ℓ runs modulo 24k2, 24k2r + k1ℓ would then run modulo 24k1k2. Write 24k r + k,ℓ t (mod 24k k ) 2 ≡ 1 2 Then t2 = (24k + k ℓ)2 (24k r)2 (mod k ) 2 1 ≡ 2 1 k2ν (mod k ), since (24r)2 ν (mod k ) ≡ 2 1 1 ≡ 1 1 Similarly t2 (k ℓ)2 (mod 24k ) ≡ 1 2 k2ν (mod 24k ), since ℓ2 ν (mod 24k ). ≡ 1 2 2 ≡ 2 2 So in order to get both conditions of summation, we need only choose 228 2 t ν (mod 24k1k2); and this can be done by the Chinese remainder theorem. So≡ = 1 3 k ℓ 1 πit/6k Bk1 (ν1)Bk2 (ν2) , − e 4 k 3 2 3 ℓ ! r t (mod 24k,k2) ! ! ≡ X This already looks very much like the ﬁrst formula though not quite. What we have in mind is to compare it with

1 k t 1 B (ν) = − eπit/6k k 4 3 3 t r t2 ν (mod 24k) ! ≡ X So ﬁnd out t 1 24k r + k ℓ 1 − = 2 1 − 3 t 3 24k2r + k1ℓ ! ! ! 26. Lecture 171

k ℓ 1 = 1 − 3 k ℓ ! 1 ! k 1 ℓ 1 = 1 − − 3 k 3 ℓ ! 1 ! ! ! 3 ℓ 1 = − , k 3 ℓ 1 ! ! ! = by the reciprocity law. So the formulas agree: Bk1 (ν1)Bk2 (ν2) Bk(ν); and we have settled the aﬀair in this case by

Theorem 1. If k2ν ν (mod k ) and k2ν ν (mod 24k ), (k, 6) = 1, then 229 2 1 ≡ 1 1 2 ≡ 2 = Bk1 (ν1)Bk2 (ν2) Bk1k2 (ν)

Case 2. This corresponds to d = d1 = 8 and d = d2 = 3. 1 k 1 k B (ν ) B (ν ) = 1 k − 2 k1 1 · k2 2 4 3 1 k 3 ! 2 ! r p 1 − eπir/2k1 eπis/3k2 . 2 r 2 (3r) ν1 (mod 8k1) ! (8r) ν2 (mod 3k2) ≡ X ≡ X 1 k 1 k k = 1 − 1 2 4 3 k 3 ! 2 ! r πi 1 s (3k2r+8k1 s) − e 6k1k2 2 3 3 (3r) ν1 (mod 8k1) ! (8r)2≡XXν (mod 3k ) ≡ 2 2 Since (k1, k2) = 1, (8k1, 3k2) = 1 andso3k2r + 8k1 s = t runs through a full system of residues modulo 24k1k2. So

1 k1 1 k 1 s = πit/(6k1k2) Bk1 (ν1)Bk2 (ν2) − − e 4 3 k2 3 2 r 3 ! ! r t ν (mod 24k1k2) ! ≡ X As before 230 t2 = (3k r + 8k s)2 (3k r)2 (3k r)2 k2ν (mod 8k ) 2 1 ≡ 2 ≡ 2 ≡ 2 1 1 t2 = (8k s)2 k2ν (mod 3k ) 1 ≡ 1 2 2 2 2 Now determine ν such that ν k2ν1 (mod 8k1) and ν k1ν2 (mod 3k2), again by the Chinese remainder theorem.≡ So t2 (mod 24≡k k ). Now ≡ 1 2 t 1 8k s 1 − = 1 − 3 t 3 3k1r ! ! ! 26. Lecture 172

k 1 s 1 = 1 − − 3 k2 3 r ! ! ! (since 8 and 1 are quadratic non-residues modulo 3). So − 1 k t 1 B (ν )B (ν ) = − eπit/6k k1 1 k2 2 4 3 3 t r t2 ν (mod 24k) ! ≡ X = Bk(ν) where ν is given. Hence Theorem 2. If k2ν ν (mod 8k ) and k2ν ν (mod 3k ), then 2 1 ≡ 1 1 2 ≡ 2 = Bk1 (ν1)Bk2 (ν2) Bk1k2 (ν)

Let us give an example of what this is good for. Calculate A10(26). Since we can reduce modulo 10, A10(26) = A10(6). ν = 1 24n = 143. − − A (26) = A (6) = B ( 143) = B ( 23) 10 10 10 − 10 − = B5(ν1)B2(ν2)

where ν1, ν2 are determined by the conditions 231 4ν 23 (mod 5)or ν 3 (mod5) 1 ≡− − 1 ≡− and 25ν 23 (mod 48)or ν 1 (mod48) 2 ≡− 2 ≡ = 3 = So A10(26) B5(3)B2(1), and these are explicitly known. Since 5 1, B (3) = 0. It is actually not necessary now to calculate B (1). − 5 2 1 πr B (1) = ( )λ − 2λ/2 sin 2 − r 2λ+1 ! where (3r)2 ν (mod 2λ+3), (3r)2 1 (mod16), ≡ ≡ or 3r 1 (mod 16), r 11 (mod 16). (there being four solutions). Then ≡ ≡ 11π 1 B2(1) = ( )( ) √2 sin = 1 √2 = 1 − − 4 × · √2

A10(26) = 0. One more thing can be established now. We have the inequalities:

λ/2 B λ (ν) 2 , | 2 |≤ 26. Lecture 173

λ B3λ (ν) 3 2 2 √3, | |≤ λ B λ (ν) 2p 2 , p > 3. | p |≤ By the multiplicative property, 232

ǫ B (ν) < C √k k ,ǫ > 0, C = Cǫ . | k ·

We see that although An(n) has ϕ(k) summands and in general all that one knows is that ϕ(k) k 1, because of strong mutual cancellations among the ≤ − 1 +ǫ roots of unity, the order is brought down to that of k 2 . This reminds us of other arithmetical sums like the Gaussian sums and the Kloosterman sums. Lecture 27

We now give a proof of the transformation formula for η(τ). η(τ) we ﬁrst in- 233 troduced by Dedekind in his commentary on a fragment on modular functions by Riemann; it is natural in the theory of elliptic functions.

πiτ ∞ 2πimτ η(τ) = e 12 (1 e ) = − Ym 1 aτ + b We want to replace τ by τ = . Actually in the whole literature there ′ cτ + d is no full account except in a paper by W.Fischer (Paciﬁc Journal of Mathe- 1 + matics, Vol. 1). We know what happens in the special cases τ and τ 1. We get the explicit form in which the root of unity appears in the− transformation formula if we put together some things from the theory of modular functions. There some discussion in Tannery-Molk; they write h(τ) instead of η(τ).(η(τ))3 V 1 8 is up to a factor 1 (o/τ). It turns out for quite other reasons that (η(τ)) can be discussed too; it has to do with the modular invariant J(τ). Dedekind did something more than what is needed here. He studied log η(τ). For Im τ > 0, η(τ) is a function in the interior of the unit circle (if we set x = e2πiτ) free from zeros and poles. So the logarithm has no branch points and is fully deﬁned without ambiguity.

πiτ ∞ log η(τ) = + log(1 e2πimτ) 12 − m 1 X− (For purely imaginary τ, the logarithms on the right side are real). The multiplicative root of unity now appear as something additive. This 234 is what Ded´ekind investigated. Recently (Mathematika, vol.1, 1954) Siegel 1 published a proof for the particular case , using logarithms. Actually Siegel − τ 174 27. Lecture 175

proves much more than the functional equation for η(τ). He proves that

1 1 τ log η( τ− ) = log η(τ) + log − 2 i We shall extend his proof to the more general case. The interesting case where a root of unity appears explicitly has not been dealt with by Siegel. We write the general modular transformation in the form

h + iz h′ + i/z τ = , τ′ = , hh′ 1 (mod k) k k ≡− We wish to prove that

h + i/z h + iz 1 log η ′ = log η + log z + πiC(h, k)(*) k k 2 ! ! where C(h, k) is a real constant. From the deﬁnition of η(τ),

h + iz π(h + iz) ∞ ∞ 1 log η = e2πimr(h+iz)/k k 12k − = = r ! Xm 1 Xr 1 πih πz ∞ ∞ 1 2πimrh/k 2πmrz/k = e e− 12k − 12k − = = r Xm 1 Xr 1 e2πimrh/k is periodic with period k; we emphasize this and write 235

m = qk + µ; µ = 1,..., k; q = 0, 1, 2,....

Then

+ k h iz πih πz ∞ ∞ 1 2πiµ rh 2π(1k+µ) rz log η = e k e− k , k 12k − 12k − = = = r ! Xq 0 Xν 1 Xr 1 and taking the summation over q inside, this becomes

πih πz ∞ ∞ 1 2πiµ rh 2πµ rz ∞ 2πqrz e k e− k e− 12k − 12k − = = r = Xµ 1 Xr 1 Xq 0 k 2πµrz/k πih πz ∞ 1 rh e− = 2πiµ k e 2πrz 12k − 12k − = = r 1 e− Xµ 1 Xr 1 − 27. Lecture 176

h′+i/z Substituting in (*), with similar expansion for η k , we have

k 2πνr kz πih π ∞ 1 2πiνr h′ e− e k 2πr/z 12k − 12kz − = = r 1 e− Xν 1 Xr 1 − k 2πµ rz 1 πih πz ∞ 1 2πiµ rh e− k = z + π , + k log iC(h k) e 2πrz 2 12k − 12k − = = r 1 e− Xµ 1 Xr 1 − Rearranging this, we get 236

k 2πνr/kz k 2πµrz/k ∞ 1 2πiνr h′ e− ∞ 1 2πiµrh/k e− k . e 2πr/z e 2πr/z = = r 1 e− − = = r 1 e− Xν 1 Xr 1 − Xµ 1 Xr 1 − π 1 πi 1 + z + (h h′) + πiC(h, k) = log z. 12k z − 12k − −2 ! We now follow Siegel’s idea to get the whole thing as a sum of residues of a certain function. Clearly there is r in it. Being integers r can be produced 1 by something like which has poles with residue 1 at every integral 1 e2πix − 2πi valued x. So let us study− a function like 1 1 e 2πµxz/k e2πiµxh/k − x 1 e2πix 1 e 2πxz − − − We may have to sum this from µ = 1 to µ = k. This should somehow be the form of the function that we wish to integrate. We do not want it in the whole plane. In fact, we can either take a wider and wider path of integration, or multiply the function by a factor and magnify it; we prefer to do the latter. 237 = + 1 We shall put xN for x, keep the path ﬁxed and take N n 2 , n integer, to avoid integral points, and then make n . The term corresponding to → ∞ 2πxz µ = k should be treated separately, as otherwise the factor e− would stop convergence. Also µh and µ should appear symmetrically for reasons which we shall see. So introduce µ µh (mod k), µ = 1, 2,..., k 1, and choose ∗ ≡ − 1 µ∗ k 1. It turns out, taking all this together, that the following thing will≤ do.≤ Write−

k 1 z 1 πNx − 1 e2πµNx/k e 2πiµNx/k = π + − Fn(x) cot h Nx cot 2πNx 2πiNx/z −4ix z = x · 1 e · 1 e− Xµ 1 − − The ﬁrst term is a consequence of the term for µ = k: 2πNxi 2πxNz 1 e e− x × 1 e2πixN × 1 e 2πxNz − − − 27. Lecture 177

The poles will not change if we write this as

2πNxi 2πxNz + 2πiNx + 2πxNz 1 e + 1 e− + 1 = 1 1 e 1 e− x 1 e2πiNx 2 1 e 2πxNz 2 x 2(1 e2πiNx) · 2(1 e 2πxNz) − ! − − ! − − − 1 = cot πxN cot hπxNz. 4xi ·

We integrate Fn(x) along a certain parallelogram P, a little diﬀerent from 238 Siegel’s. P has vertices at z, i (since τ> 0, z > 0). Then ± ± Jm Re 1 Fn(x)dx = (Residues). 2πi p Z X

We then let n . →∞ The poles of Fn(x) are indicated by the denominators and the cotangent factors. These are rz ir x = 0, x = , x = , r integer. − N N x = 0 is a triple pole for the ﬁrst summand.

1 πNx 1 1 z cot hπNx cot = − 4ix z −4ix · πNx πNx (πNx)2 (πNx/z)2 1 + + 1 + 3 · × − 3 ··· ( ) ( ) 27. Lecture 178

Residue for this term at x = 0 239 iz 1 π2N2 = π2N2 4π2N2 · 3 − z ! i 1 = z . 12 − z ! which had been foreshadowed already. Lecture 28

We had 240

k 1 z 1 πNx − 1 e2πµNx/k e 2πiµ∗Nx/k = π + − , Fn(x) cot h Nx cot 2πNx 2πiNx −4ix z = x · 1 e × 1 e− /z Xµ 1 − − N = n + 1 , n integer > 0, µ hµ (mod k) and 1 µ k 1. At the triple 2 ∗ ≡ ≤ ∗ ≤ − = = 1 1 pole x 0 the residue from the ﬁrst summand 12i z z - Let us ﬁnd the residues from the more interesting pieces of the sum.− The− general term on the right has in the neighbourhood of x = 0 the expansion

1 2πµNx (2πµN /k)2 1 + + 2 + x k 2! ··· ( ) 1 1 2πNx (2πNx)2 − − 1 + + + × 2πNx 2 6 ··· ( ) 2πiµ Nx (2πµ Nx/kz)2 1 ∗ ∗ + × − kz − 2 ··· ( ) 1 1 2πiNx (2πNx/z)2 − 1 + × 2πiNx/z − 2z − 6 ··· ( ) z 2 = + 2πµNx + 1 2πµNx 2− 2 3 1 4π iN x  k 2 k −···   !   2  2πNx (2πNx) 2 1  + + + ( ) + × − 2 6 ··· ··· ··· ( ! ) 2πiµ Nx 1 2πµ Nx 2 1 ∗ ∗ + × − kz − 2 kz ···  !       179  28. Lecture 180

2πiNz (2πN2/z)2 1 + + + × z 2 ··· ( ) 241 1 Fishing out the term in , the residue at x = 0 from this summand becomes x

iz 1 2πµN 2 1 1 2πµ N 2 1 2πN 2 N + π 2 ∗ πµ π 2 2 (2 N) 2 N 4π N 2 k 12 − 2 kz − 12 z − k  ! ! !  2 2 2 2 2 2 2 2 2  2 N i 2π iµN 2π iµ∗N π iN 2π µ∗N  4π µµ∗ + + + − k2z kz kz − z kz2 ) z 2 = i 2 2 + 1 2µ + i 2µ∗ 1 + 2µ∗ 2 k − 2 4 µ 3 − k 4z  k − 3 k  ( )    i 4iµµ 2iµ 2iµ  + ∗ +  + ∗ i  4 − k2 k k − ( ) µ2 µ 1 1 µ 2 µ 1 µ 1 µ 1 = iz + + ∗ ∗ + + ∗ (*) 2k2 − 2k 12 iz 2k2 − 2k 12 k − 2 k − 2 ( )   ! !   = = We have to sum this up from µ 1 to µ k 1. Let us prepare a few 242 things. − Let us remark that

k 1 k 1 − (k 1)k − (k 1)k(2k 1) µ = − ; µ2 = − − = 2 = 6 Xµ 1 Xµ 1

Also if µ runs through a full system of residues, so would µ∗ because = µ∗ µ∗ hµ ﬀ (h, k) 1. Further 0 < k < 1, and k and k di er only by an integer, so µ∗ = hµ hµ = that k k k . Hence summing up the last expression ( ) from µ 1 to µ = k 1, we− have ∗ − h i (k 1)(2k 1) k 1 k 1 ız − − − + − 12k − 4 12 ( ) k 1 1 (k 1)(2k 1) k 1 k 1 − µ 1 hµ hµ 1 + − − − + − + iz 12k − 4 12 = k − 2 k − k − 2 ( ) Xµ 1 ! " # ! 2k 1 1 1 = (k 1) − iz + + s(h, k) − 12k − 6 iz ! ! 28. Lecture 181

where s(h, k) stands for the arithmetical sum

k 1 − µ 1 hµ hµ 1

= k − 2 k − k − 2 Xµ 1 ! " # ! which appears here very simply as a sum of residues. The last expression becomes k 1 1 − iz + + s(h, k) − 12k iz ! So the total residue at x = 0 is 243 1 1 k 1 1 1 1 iz + − iz + + s(h, k) = iz + + s(h, k) 12 iz − 12k iz 12k iz ! ! ! = ir , Next, we consider the simple poles of Fn(x) at the points x N (r 0). The coth factor is periodic and so the residue at any of these poles is the same 1 = ir , as that at the origin, which is π . Hence the residue of Fn(x) at x N (r 0) becomes k 1 2πµ∗r/kz N 1 πir − N 1 2πiµ r e cot + e k − 2πr/z 4r · πN z = ir 2πN 1 e Xµ 1 − (There is a very interesting juxtaposition of an arithmeticalterm and a function theoretic term in the last part; this gets reversed for the next set of poles)

k 1 2πµ r/kz 1 πr 1 − 1 µr e ∗ = 2πi k coth e 2πr/z 4πir z − 2πi = r 1 e Xµ 1 − r x remains between i on the imaginary axis. So N < 1; so we need consider only r = 1, 2,...,± n. Again, ± ± ±

ey + e y 2e y coth y = − = 1 + − ey e y ey e y − − − − 2e 2y = 1 + − 1 e 2y − − 1 coth y is an odd function so that y cot hy is even. Hence summing up over all 244 the poles corresponding to r = 1,..., n, we get the sum of the residues ± ±

n 2πr/z k 1 n 2πµ r/kz 1 1 2e− 1 − 1 e− ∗ = + + 2πih′ µ∗r/k 1 2πr/z e 2πr/z 2πi = r 1 e− 2πi = = r 1 e− Xr 1 ( − ) µX∗ 1 Xr 1 − 28. Lecture 182

k 1 n z 1 − 1 e2πµ∗r/k 2πih′(k µ∗)r/k , e − 2πr/z − 2πi = = r 1 e µX∗ 1 Xr 1 − where we have made use of the fact that hh 1 (mod k), so h µ hh µ ′ ≡ − ′ ∗ ≡ ′ ≡ µ (mod k), or µ h′µ∗ (mod k). In the last sum replace µ∗ by k µ∗; then −the previous sum is≡− duplicated and we get −

n 2πr/z k 1 n 2πµ r/kz 1 1 2e− 1 − 1 e− ∗ + + 2πih′ µ∗r/k 1 2πr/z e 2πr/z 2πi = r  1 e− πi = = r 1 e−  r 1  − µ∗ 1 r 1 −  X  X X  n k n 2πνn/kz  1 1 1 1 e−  = + 2πih′νr/k  e 2πr/z 2πi = r πi = = r 1 e− Xr 1 Xν 1 Xr 1 − This accounts for all the poles on the imaginary axis (except the origin 245 which has been considered separately before). = rz , Finally we have poles x N (e 0) on the other diagonal of the parallelogram. The same calculation goes through verbatim and we get the sum of the residues at these poles to be

n k n 2πνrz/k i 1 + i 1 2πihνr/k e− e 2πrz 2π = r π = = r 1 e− Xr 1 Xν 1 Xr 1 − Lecture 29

We had 246

k 1 z 1 πNx − 1 e2πµNx/k e 2πiµ∗Nx/k = π + − Fn(x) cot h Nx cot 2πNx 2πNx/z −4ix z = x 1 e 1 e− Xµ 1 − − The residue at x = 0 is 1 1 iz + + s(h, k), 12k iz ! s(h, k), which will interest us for some time, being

k 1 − µ 1 hµ hµ 1 . k − 2 k − k − 2 µ=1 X ! " # ! = ir , The residues at the points x N (r 0) amount to

n k n 2πνr/kz 1 1 1 1 r e− + 2πih′ν k e 2πr/z ; 2πi = r πi = = r 1 e− Xr 1 Xν 1 Xr 1 − = zr , and the residues at the points x N (r 0)

n k n 2πµrz/k i 1 i 1 r e− + 2πihµ k e 2πrz 2π = r π = = r 1 e− Xr 1 Xµ 1 Xr 1 − n 1 When we add up, the sums r=1 r , the disagreeable ones which would have 247 gone to inﬁnity, fortunately destroy each other; so the sum of the residues of P Fn(x) at all its poles becomes

183 29. Lecture 184

k n 2πνr/kz 1 1 1 1 e− z + , + 2πih′νr/k s(h k) e 2πr/z 12ki z − πi = = r 1 e− ! Xν 1 Xr 1 − k n 2πµrz/h 1 1 2πihµr/k e− e 2πrz − πi = = r 1 e− Xµ 1 Xr 1 − We had prepared in advance what we were going to obtain. s(h, k) is what we had called C(h, k) + (h h′)/12k. We have to prove that the sum of the −= h h′ 1 residues above, with C(h, k) s(h, k) 12−k , is equal to 2πi log z, as n . But there is one difference. The sums− we have earlier were− sums from →r = ∞1 to r = ; whereas here they are sums from r = 1 to r = n. But this does not ∞ z matter as convergence is guaranteed since we have an exponential factor e− with z > 0. We have to see what becomes of our sum when we evaluate it Re in another way. We have to consider lim Fn(x)dx. So in effect we have to n p →∞ prove that R 1 1 lim Fn(x)dx = log z. n 2πi 2πi →∞ p − Z 248 Now this is a question of direct computation. Let us look at the path of integration. Fn(x) will be seen to have simple limits on the sides of the parallelogram. We consider xFn(x) broken into pieces. Take the first piece 1 πNx cot hπNx cot 4i z On the side from x = i to x = z,

x = ρ + σz; ρ, σ 0, ρ + σ = 1. i ≤ Actually we take only ρ,σ > 0; we shall exclude the points i and z themselves. Then this becomes 1 eπN(ρi+σz) + e πN(ρi+σz) eπiN(ρi+σz)/z + e πiN(ρi+σz)/z − i − −4i eπN(ρi+σz) e πN(ρi+σz) × × eπiN(ρi+σz)/z e πiN(ρi+σz)/z − − − − πNσz Nπσz The size of the first factor is determined by the terms e and e− in the numerator; the first term becomes big and the other goes to zero as N (σ> 0 and z > 0). So we divide by the first term. Similarly for the → ∞ Re 29. Lecture 185

second factor. We therefore get

ρ 2πN(ρi+σz) 2πN + σ 1 1 + e− e− z i + 1 + ρ −4 1 e 2πN(ρi σz) · 2πN + σ − − e− z i 1 − 249 As N the exponential factors go to zero; so the whole expression 1 → ∞ tends to 4 . It will further remain on its way bounded, because the numerators in either factor are at most equal to 2, while the denominators remain away from zero by a fixed amount, as we shall be showing in a moment - and for this = + 1 it is essential to have N n 2 . Since the functions concerned are even functions, what was good here would also be good on the apposite side, from x = i to x = z. So on 1 − − this side also the expression will tend to 4 . We cannot say uniformly; indeed if σ = 0, here is no convergence in the first factor, and if ρ = 0 none in the second factor, though there is boundedness: the thing would oscillate finitely. Now take the other pieces of xFn(x) on the same sides of ρ. We have to consider N N 2πµ (ρi+σz) 2πiµ∗ (ρi+σz) e k e− kz 2πN(ρi+σz) 2πi N (ρi+σz) 1 e × 1 e− z − − Remember, what is now important, that 0 <µ< k, but neither 0 nor k. The denominator in the first factors goes more strongly to infinity as N µ → ∞ than the numerator because k is a proper fraction; so too in the second factor because µ> 1. So the whole function tends to zero. Hence on these two sides xF (x) 1 . n → 4 Now consider the other two sides; it looks different here and has got to be 250 inspected. On the side from x = i to x = z, x = ρi + σz; σ,ρ > 0, σ + ρ = 1, − − and the first part of xFn(x) is

1 πNx 1 eπN( ρi+σz) + e πN( ρi+σz) π = − − − cot h Nx cot πN( ρi+ σz) − πN( ρi+σ z) −4i z −4 e − − e− − − − ρi ρi πiN − +σ πiN − + σ e z + e− z − ρ ρi πiN − i+σ πiN +σ × e z e− − z + − ρi + + 2πN( ρi σz) + 2πiN z σ = 1 1 e− − 1 e− − 2πN( ρi+σz) ρ i −4 1 e −× 2πiN − +σ − − − 1 e− − z − Let N . Assuming that the denominator is going to behave decently, →1 ∞ this goes to 4 . The other pieces go to zero for the same reason as before. And all this is good− for the opposite side too. 29. Lecture 186

We now have got to show that the convergenceit nice and the denominators do not make any fuse. This we can clarify in the following way. Consider the 2πN(ρi+σz) denominator 1 e− . Diﬃculties− will arise if the exponent comes close to an even multiple of πi. So we should see that it stays safely away from these points.

251

And actually it stays away from the danger spots by the same distance, for 251 the exponent is 2N(πiρ + πzσ) i.e., a point on the segment joining (2r + 1)πi and (2r + 1)πz.− Since ez is periodic there is a minimal amount by which it stays away from 1. The second denominator looks a little diﬀerent. We have π z instead of πz. But we have only to turn the whole thing around. We see how = + 1 = + 1 = 1 essential it was to take N n 2 (2n 1) 2 on odd multiple of 2 . So the convergence is nice, but not uniform. We can nevertheless say that 1 xFn(x) 4 boundedly on the sides of ρ except for the vertices where it does not converge→± but oscillates ﬁnitely. But bounded convergence is enough for 1 interchanging integration and summation. Fn(x) 4x and the x does not ruin anything because it stays away from zero everywhere→ ± on ρ. Hence 1 lim Fn(x)dx n 2πi →∞ Zp exists and we have 1 1 1 lim Fn(x)dx = dx n 2πi 2πi ±4x →∞ Zp Zp i z i z = 1 dx − dx + − dx dx 2πi  4x − i 4x z 4x − i 4x  Zz Z Z− Z−     z z  1  i dx dx i dx dx  =  +  8πi z x − i x z x − i x (Z Z− Z Z− ) 29. Lecture 187

i z = 1 dx dx 4πi x − x (Zz Zi ) z is in the positive half-plane; we can take the principal branch of the logarithm, 252 so that we get on integration, since log i is completely determined, 1 π πi 1 log z log z + = log z 4πi 2 − − 2 −2πi

So we have proved the foreseen formula with the particular substitution h h C(h, k) = s(h, k) − ′ : − 12k h + i/z h + iz 1 h h log η ′ = log η + log z + πis(h, k) + πi ′ − , k k 2 12k ! ! which is the complete formula in all its details. The mysterious s(h, k) enjoys 253 certain properties. It has the group properties of the modular group behind it and so must participate in them. Lecture 30

Last time we had the formula of transformation of log η in the following shape: 254

h′ + i/z h + iz 1 πi log η = log η + log z + (h′ h) + πis(h, k), k k 2 12k − ! ! where s(h, k) is the Dedekind sum, which, by direct computation of residues, was seen to be k 1 − µ 1 hµ hµ 1 . = k − 2 k − k − 2 Xµ 0 ! " # ! We use the abbreviation: for real x,

x [x] 1 , if x is not an integer, ((x)) = − − 2   0 , if x is an integer.   Then  k µ hµ s(h, k) = . = k k Xµ 1 !! Now ((x)) is an odd function; for x integer, trivially (( x)) = ((x)), and for x not an integer, − − 1 (( x)) = x [ x] − − − − − 2 1 = x + [x] + 1 , since [ x] = [x] 1, − − 2 − − − = ((x)). − ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗ 188 30. Lecture 189

((x)) is the familiar function whose graph is as indicated. 255

We now prove that k µ = 0 = k Xµ 1 Because of periodicity we can write

k µ = µ = k k Xµ 1 µ Xmod k = µ k µ Xmod k = µ − k µ Xmod k k µ ∴ = 0 = k Xµ 1 We can also write s(h, k) in the form

k µ 1 hµ s(h, k) = = k − 2 k Xµ 1 ! !! k µ hµ 1 k hµ = , = k k − 2 = k muX1 !! Xµ 1 !! and since hµ also runs through a full system of residues mod k when µ does 256 so, as (h, k) = 1, the second sum is zero, and we can therefore write

k µ hµ s(h, k) = = k k Xµ 1 !! 30. Lecture 190

Let us now rewrite this in a form in which the modular substitution comes into play h′ + i/z h + iz τ′ = , τ = ; k k so kτ h = iz, and − h′ 1/(kτ h) h′kτ hh′ 1 τ′ = − − = − − k k(kτ h) − h′τ (hh′ + 1)/k = − kτ h − hh′+1 ( k is necessarily integral for hh′ 1 mod k). So the modular substitution is ≡− + h hh′ 1 a b ′ − k = , c > 0.  k h  c d  −      η  The transformation formula for log  now reads aτ + b 1 cτ + d πi log η = log η(τ) + log + (a + d) πis(d, c), cτ + d 2 i 12c − ! since s( d, c) = s(d, c). − −− Let us take in particular 257

a b 0 1 = − ; c d 1 0          then we obtain     1 1 τ log η = log η(τ) + log , τ 2 i ! the special case discussed by Siegel. Let us now make two substitutions in succession: + = aτ′ b = 1 τ′′ + , τ′ . cτ′ d − τ Then a/τ + b bτ a τ′′ = − = − c/τ + d dτ c − − We suppose c > 0, d > 0; (c, d) = 1. Then

1 cτ′ + d πi log η(τ′′) = log η(τ′) + log + (a + d) πis(d, c); 2 i 12c − 30. Lecture 191

1 dτ c πi log η(τ′′) = log η(τ) + log − + (b c) πis( c, d). 2 i 12d − − − Sub tracting, and observing that 1 τ log η(τ′) log ηn(τ) = log , − 2 i we have 258 1 τ 1 cτ + d 1 dτ c 0 = log + log ′ log − 2 i 2 i − 2 i πi a + d b c + − πi(s(d, c) s(c, d)) 12 c − d − − ! The sum of the logarithms on the right side is determinate only up to a multiple of 2πi: τ cτ + d dτ c τ ( c/τ + d)/i log + log ′ log − = log − + 2πik 1 i − i i (dτ c)/i − 1 = log + 2πik i ! πi = + 2πik − 2 Now each logarithm above has an imaginary part which is strictly less than π 2 in absolute value; so τ cτ + d dτ c 3π Im log + log ′ log − < i i − i 2 ( )

So the only admissible value of k is zero.

Hence we have 259 πi πi a + d b c 0 = + − πi(s(d, c) + s(c, d)) , − 4 12 c − d − ! or since ad bc = 1, − 1 1 d c 1 s(d, c) + s(c, d) = + + + . −4 12 c d cd ! This is the reciprocity law for Dedekind sums. It is a purely arithmetical formula for which I have given several proofs; here I reproduce the proof that I gave originally, by lattice-point enumeration. 30. Lecture 192

We have to prove that

k 1 h 1 − µ hµ hµ 1 + − ν kν hν 1 = k k − k − 2 = h h − k − 2 Xµ 1 ( " # ) Xµ 1 ( " # ) 1 1 h k 1 = + + + , −4 12 k h hk or ! k 1 k 1 h 1 h 1 k 1 h 1 h − 1 − k − 1 − 1 − hµ 1 − kν µ2 µ + ν2 ν µ ν k2 − 2k h2 − 2h − k k − h h µ 1 µ 1 ν 1 ν=1 µ=1 ν=1 X− X− X− X X " # X " # 1 1 h h 1 = + + + ; −4 12 k h hk ! or 260 h2(k 1)(2k 1) h k(k 1) k2(h 1)(2h 1) k h(h 1) − − − + − − − 6 − 2 2 6 − 2 2 k 1 h 1 − hµ − kν h µ k ν − = k − = h Xµ 1 " # Xν 1 " # 3hk + h2 + k2 + 1 = − 12 k 1 h 1 − hµ − kν i.e., 12h µ + 12k ν = k = h Xµ 1 " # Xν 1 " # = h(k 1)(2h(2k 1) 3k) + k(h 1)(2k(2h 1) 3h) + 3hk h2 k2 1 − − − − − − − − − = 8h2k2 9h2k 9hk2 + h2 + k2 + 9hk 1 − − − = (h 1)(k 1)(8hk h k 1) − − − − − So the whole thing is equivalent to proving that

k 1 h 1 − hµ − kν 12h µ + 12k ν = (h 1)(k 1)(8hk h k 1). = k = h − − − − − Xµ 1 " # Xν 1 " # This reduces to something that looks familiar; indeed the square brackets appear in lattice-point enumeration. Here (h, k) = 1, but in a paper with White- man I have also discussed the case where h, k are not coprime. Enumerating by rows and columns parallel to the µ and ν axes, the num- 261 ber of lattice-points in the integer a the rectangle − − 30. Lecture 193

with sides of length k, h along the axes of µ and ν respectively is seen to be (h 1)(k 1). This can beenumeratedin anotherway also. Thenumberof lattice points− in− the interior, with abscissa µ and lying below the diagonal through the k 1 hµ − hµ origin is the full integer in k . So we have k lattice points below the µ=1 h 1 h i − kν P = diagonal. Similarly there are h points abovethe diagonal. Since (h, k) 1 ν=1 there are no points on the diagonal.P h i Hence

k 1 h 1 − hµ − kν (h 1)(k 1) = + − − = k = h Xµ 1 " # Xν 1 " # In out case we have quadratic summands; but something which goes so well here in the plane should go well in space also. Lecture 31

We want to prove directly the reciprocity formula 262

1 1 h k 1 s(h, k) + s(k, h) = + + + −4 12 k h hk ! k µ hµ with s(h, k) = = k k Xµ 1 !! The reciprocity formula is equivalent to proving that

k 1 h 1 − hµ − kν 12h µ + 12k ν = (h 1)(k 1)(8hk h k 1) = k = h − − − − − Xµ 1 " # Xν 1 " # We made a little digression and spoke of similar sums which occur in lattice-point summations:

k 1 h 1 − hµ − hν + = (h 1)(k 1) = k = h − − Xµ 1 " # Xν 1 " # h k If we use a rectangle of sides 2 , 2 , (h, k odd) we obtain

k 1 h 1 −2 hµ −2 hν 1 + = (h 1)(k 1). = k = h 4 − − Xµ 1 " # Xν 1 " # This is made use of the theory of quadratic residues. The summands in our case are ‘quadratic’ in µ and ν.

194 31. Lecture 195

Consider the rectangular parallelopiped with three concurrent edges along 263 the axes of µ, ν and ρ, the lengthsof these edgesbeing h, k, hk respectively. Dis- sect the parallelopiped into three pyramids having a common apex at the origin and having for bases the three rectangular faces which do not pass through the origin, viz. ABCD, BCFE and CDGF. We now compute the number of lattice points in each pyramid. Take for example the pyramid O(BEFC). Consider a section parallel to the (ρ, ν)-plane at a distance µ along the µ-axis. The lat- h tice points lie in such sheets. The edges of this section are hµ and µ k . The number of lattice points on this sheet (including possibly those on the edges) is k 1 µh = − µh hµ k . So for the whole pyramid the number hµ k . For the pyramid µ=1 h i h 1 h i − Pνk O(ABCD), the one facing us, the number is kν h ν=1 Of course are some points on the commonP edge.h Finallyi there is a pyramid 264 of exceptional sort which lies upside down. Consider a section at a height h parallel to the (µ, ν) plane the numberr of lattice points on and inside this pyramid is seen to be hk 1 − ρ ρ . = h k Xρ 1 So altogether we have

k 1 h 1 hk 1 − µh − νk − ρ ρ hµ + kν + = k = k = h k Xµ 1 " # Xν 1 " # Xρ 1 31. Lecture 196

points, including some points which have been counted twice over. But the number of lattice points inside: the parallelopipedis equal to (h 1)(k 1)(hk 1). Hence making a correction for the lattice points on the cleaving− − surfaces− through the edges CF and CD which have been counted twice (the surface along BC has no points on it because (h, k) = 1), we have

k 1 h 1 hk 1 − µh − νk − ρ ρ hµ + kν + = k = h = k h Xµ 1 " # Xν 1 " # Xρ 1 = (h 1)(k 1)(hk 1) + (h 1)(k 1) − − − − − = hk(h 1)(k 1) − − Now write

hk 1 − ρ ρ S = = h k Xρ 1 ρ ρ 1 ρ ρ ρ = , if h ∤ ρ; , if h ∤ ρ. h h − 2 − h h − h So 265 hk 1 − ρ 1 ρ ρ 1 ρ S = = h − 2 − h 2 − 2 − k Xρ 1 ( )( ) With some correction. Indeed h ρ, k ρ do not happen together: Let ρ = hσ, ρ = kτ. In the ﬁrst case. i.e., h| ρ, we| have to correct the above by an amount | k 1 − 1 hσ 1 hσ , = 2 k − 2 − k Xσ 1 ( !!) and in the second case, k ρ, by | h 1 − 1 kτ 1 kτ

= 2 h − 2 − h Xτ 1 ( !!) So

hk 1 hk hk − ρ 1 ρ 1 ρ ρ 1 ρ ρ 1 S = = h − 2 k − 2 − = h k − 2 − = k h − 2 Xρ 1 ( )( ) Xρ 1 ! Xρ 1 ! hk k 1 h 1 + ρ ρ + 1 − hσ 1 + 1 − 1 kτ 1 = h k 2 = k − 2 2 = 2 h − 2 Xρ 1 Xσ 1 ( ) Xτ 1 ( ) 31. Lecture 197

µ = Since k 0, thisbecomes µ mod k P hk 1 hk hk − ρ2 1 ρ ρ 1 1 ρ 1 ρ S = + + ρ ρ = hk − 2 h k 4 − 4 = k − h = k Xρ 1 ( ) Xρ 1 Xρ 1 hk ρ ρ 1 h(k 1) k 1 1 k(h 1) h 1 + + − − + − − = h k 2 2 − 2 2 2 − 2 Xρ 1 ! ! we use the periodicity in the non-elementary pieces; so write 266

ρ = hr + s; r = 0, 1,..., k 1; s = 1,..., h. − hk k 1 h ρ − hr + s ρ = (hr + s) = h = = h Xρ 1 Xr 0 Xs 1 !! k 1 h k 1 h − s − s = hr + s = = h = = h Xr 0 Xs 1 Xr 0 Xs 1 h s = k s = h Xs 1 (since the ﬁrst sum is zero, as we see by summing over s ﬁrst)

h 1 − s 1 = k s = h − 2 Xs 1 ! (h 1)(2h 1) 1 = k − − h(h 1) 6 − 4 − ( ) k(h 1)(h 2) = − − 12 Similarly 267 hk ρ h(k 1)(k 2) ρ = − − = k 12 Xρ 1 next, consider ik ρ ρ

= h k Xρ 1 Write ρ = hα + kβ; when α, β run through complete systems of residues modulo h, k respectively, hα + kβ runs through a complete system modulo hk, 31. Lecture 198

by the Chinese remainder theorem. Then

hk + + ρ ρ = hα kβ hα kβ = h k h k Xρ 1 α Xmod k β Xmod h !! !! = kβ hα h k α Xmod k β Xmod h !! !! = hα kβ k h α Xmod k !! β Xmod h !! = 0 since each sum is separately zero. Hence 1 1 1 S = (hk 1)(2hk 1) (hk 1)(k + h) + (hk 1) 6 − − − 4 − 4 − 1 1 1 (h 1)(h 2) (k 1)(k 2) + (k 1)(h 1) − 12 − − − 12 − − 2 − − 1 1 = (hk 1)(4hk 3h 3k + 1) (h 1)(h 2) 12 − − − − 12 − − 1 1 (k 1)(k 2) + (k 1)(h 1) − 12 − − 2 − − 1 = (h 1)(k 1)(4hk + h + k + 1) 12 − − Thus 268

k 1 h 1 − hµ − kν 1 h µ + k ν + (h 1)(k 1)(4hk + h + k + 1) = k = h 12 − − Xµ 1 " # Xν 1 " # = (h 1)(k 1)hk − − k 1 h 1 − hµ − kµ ∴ 12h µ + 12k ν = k = h Xµ 1 " # Xν 1 " # = (h 1)(k 1)(8hk h k 1) − − − − − We make some elementary remarks about quadratic residues. The reciprocity formula gives, on multiplication by 12h2k.

12h2ks(h, k) + 12h2ks(k, h) == 3h2k + h3 + k2h − Look at the denominator of s(h, k). At worst it can have for factors 2 and 269 31. Lecture 199

k2. So2k2 s(h, k) is integral. 2h2 s(k, h) is also integral. 12h2ks(h, k) h3 + k2 + h (mod 3k) ≡ h(h2 + 1) (mod k), ≡ and since h2 cannot help to make an integer of the left side, 12hks(h, k) h2 + 1 (mod k). ≡ Sp 12ks(h, k) is an integer. The highest possible denominator for s(h, k) is (2k2, 12k) = 2k(k, 6). So the denominator which at ﬁrst glance could conceiv- ably be as big as 2k2 is actually at most only 2k(k, 6). This is achieved, for instance, in s(1, 3) = 1/18, where 6(6, 3) = 18. In fact s(1, 3) can be computed from the reciprocity formula: 1 1 1 3 1 s(1, 3) + s(3, 1) = + + + s(3, 1) = 0 −4 12 3 1 3 ! = 1 since an integer is involved and so s(1, 3) 18 . In general, 1 1 1 k 1 s(1, k) = + + + −4 12 k 1 k ! (k 1)(k 2) = − − 12k s(2, k) is also easily obtained. k is odd; so we have 1 1 2 k 1 s(2, k) + s(1, 2) = + + + −4 12 k 2 2k ! and as s(1, 2) = 0 (by direct computation), we get 270 (k 1)(k 5) s(2, k) = − − 24k Let us calculate s(5, 27). 1 12 + 52 + 272 s(5, 27) + s(27, 5) = + −4 12 5 27 × × 1 12 + 22 + 52 s(2, 5) + s(5, 2) = + −4 12 2 5 × × s(5, 2) = 0 = s(1, 2), and on sub traction, s(5, 27) = 35/(6 27); andweknowthat × the denominator could be at most 2.27(27, 6) = 6 27. × Lecture 32

We shall study a fewmore propertiesof Dedekindsums. We had the reciprocity 271 law 1 1 h k 1 s(h, k) + s(k, h) = + + + . −4 12 k h hk ! From this we deduced as a consequence

12hk s(h, k) h2 + 1 (mod k)(*) ≡ Now when do the Dedekind sums vanish? Let us write s(h, k) in the more ﬂexible form: µ µh s(h, k) = k k µ Xmod k !! Let hh∗ 1 (mod k). Since (h∗, k) = 1, h∗µ runs through a full residue system modulo≡ k, and so

µh µhh s(h, k) = ∗ ∗ k k µ Xmod k !! !! = µ µh∗ k k µ Xmod k !! = s(h∗, k)

a b This is of some signiﬁcance. We came to s from the substitution c d and since ad 1 (mod c), s(d, c) = s(a, c). hh 1 (mod k), and ≡ ′ ≡− µ µh s(h, k) = k k µ Xmod k !! 200 32. Lecture 201

= h′µ µhh′ k k µ Xmod k !! !! = µ h′µ − k k µ Xmod k !! = s(h′, k) − 2 2 When h = h′ i.e., h (mod k) (cg. 2 (mod 5)) 272 ≡− ≡− s(h, k) = s(h, k) − or s(h, k) = 0 if h2 1 (mod k) ≡− (in particular if h2 + 1 = k. I have a conjecture that s(h, k) 0 if h2 < k). In fact we can say more. We have the ≥ Theorem. 12 s(h, k) is en integer only for h2 1( (mod k)) and is then equal to zero. ≡− For assume that 12s(h, k) = integer; this implies, because of (*), that 0 h2 + 1 (mod k) ≡ In such cases, therefore, we can make a direct statement about the value of s(h, k) without going through the rigmarole of the Euclidean algorithm. Thus s(2, 5) = 0, s(5, 26) = 0. In a recent issue of the Duke Mathematical Journal (1954), I gave a generalisation of the reciprocity formula for Dedekind sums. It takes into account three summands. The formula is very elegant and throws some light on the reciprocity relation itself. We quote it without proof.

Theorem . If a, b, c are pairwise coprime and aa∗ 1 (mod bc), bb∗ 1 (mod ca), cc 1 (mod ab), then ≡ ≡ ∗ ≡

T s(bc∗, a) + s(ca∗, b) + s(ab∗, c) ≡ = 1 + 1 a + b + c −4 2 bc ca ab ! 273 The proof is by an algebraic method due to R´edel. The formula is very gratifying as a generalisation of the reciprocity formula is which latter there is some non-homogeneity. Put c = 1; then c∗ = 1, and s(ab∗, c) = 0; sowe getthe 1 + 2 + 2 + 2 reciprocity formula. The right side above is 12 abc 3abc a b c . Hence T = 0 if and only if a2 + b2 + c2 = 3abc. This combination− of three integers plays some role the theory of quadratic forms; it is called a Markoﬀ triple. It has reappeared in literature in connection with the geometryof numbers. It has 32. Lecture 202

to do with the existence if certain quadratic forms with minimum values close to zerofor integers. 1, 1, 2 is a Markoff triple. If we keep two of them fixed, for the third we get a quadratic equation of which one root we know to be rational. So the other root is rational too. For instance if a, b = 1 are fixed, we have c2 3c + 2 = 0or(c 1)(c 2) = 0; – the triples pre 1, 1, 1 and 1,1,2. If we take− the triple a, 1, 2,− then a2−+ 5 = 6a or a = 1, 5; we have the triples b, 1, 2; 1, 1, 2. T = 0 only if a, b, c being to a Markoff triple. For such a triple,

b2 + c2 0 (mod a), c2 + a2 0 (mod b), a2 + b2 0 (mod c) ≡ 2 2 ≡ 2 ≡ So b c (mod a), or (c∗b) 1 (mod a), etc. ≡− ≡−

Then s(bc∗, a) = 0, and each summand in T is zero. Dedekind sums have something to do with Farey fractions. Let us suppose a b = that c d 1, c, d > 0.

1 1 c d 1 s(c, d) + s(d, c) = + + + −4 12 d c cd ! cb 1 (mod d) and ad 1 (mod c) ≡− ≡ so s(c, d) = s( b, d) and s(d, c) = s(a, c). − − 1 c d 1 So s(a, c) s(b, d) = 4 + + + − − 12 d c cd ! h h h h Now if 1 , 2 the adjacent Farey Fractions, then 1 2 = 1 274 k1 k2 k1 k2 so −

1 1 k1 k2 1 s(h , k ) s(h , k ) = + + 1 1 − 2 2 4 − 12 k k k k 2 1 1 2 ! Write the left side as s h1 s h2 . k1 − k2 Suppose h2 is fixed. Let us look at all possible adjacent fractions h1 . They k2 k1 + are obtainable by forming mediants; replace h1 successively by h1 λh2 . Make k1 k1+λk2 k2 1 k1 h1 h2 k1 larger and larger. Then and 0. So . Thus s s k1 k1k2 k2 k1 k2 h → → ∞ − goes unboundedly by , and so s 1 . Therefore only on the left side −∞ k1 → −∞ of h2 can we get a sequence of rational fractions for which the Dedekind sums k2 tend to . We−∞ now give another proof of the reciprocity law, by the method of finite µ Fourier series, k is a number- theoretic periodic function. It has a finite Fourier expansion: k µ 2πi j µ = c je k k = Xj 1 32. Lecture 203

In fact this is always solvable for c1, c2,..., ck. For writing down µ = 275 1, 2, 3,..., k in succession, we have a system of k linear equations whose determinant is a Vandermonde determinant which is non-zero since the roots of unity are diﬀerent. We have

k k k ℓ ( j ℓ) µ 2πiµ 2πiµ − e− k = c j e k = k = = Xµ 1 Xj 1 Xµ 1 = kcl, k 1 µ 2πiµ ℓ i.e., cl = e− k k = k Xµ 1 This was done by Eisenstein, We can also write

k 1 1 − µ 1 2πiµ l cl = e− k k = k − 2 Xµ 1 ! k 1 k 1 − l 1 2πiµ 2−k , if k l; = µe− k | k2 −  µ=1  1 , if k ∤ l. X  2k  So if k l, then  | 1 k(k 1) k 1 cℓ = − − = 0. k2 2 − 2k In particular ck = 0. If k ∤ l, then writing

k 1 − 2πiµ ℓ S = µe− k , = Xµ 1 k 1 + 2πi ℓ − 2πi µ 1 ℓ Se− k = µe− k = Xµ 1 k 2πiν ℓ = (ν 1)e− k = − Xν 2 k 2πi l 2πiν l 2πi l = S e− k + k e− k + e− k − − = Xν 1 32. Lecture 204

So 276 k S = e 2πiℓ/k 1 − − Hence, if k ∤ l, then 1 1 c = − + ℓ k(1 e2πiℓ/k) 2k − + 2πiℓ/k = 2 1 e− − − 2πiℓ/k 2k(1 e− ) − ℓ + 2πi k = 1 1 e− 2πi ℓ −2k · 1 e− k i πℓ− = cot 2k k So we have what is essentially Eisenstein’s formula:

k 1 µ i − πℓ 2πi j µ = cot e k k 2k = k Xj 1 µ This is an explicit formula for k as a ﬁnite Fourier series. We utilise it for Dedekind sums. µ hµ s(h, k) = k k µ Xmod k !! k 1 k 1 1 − π j µ − πℓ µ = 2πi j k 2πiℓh k 2 cot e cot e −4k = k × = k µ Xmod k Xj 1 Xℓ 1 k 1 k 1 1 − − π j πl µ + = 2πi k ( j hℓ) 2 cot cot e −4k = = k k Xj 1 Xℓ 1 µ Xmod k k 1 1 − πℓ πhℓ = cot cot − , −4k = k k Xℓ 1 since in the summation with respect to µ only those terms remain for which 277 j + hℓ (mod k). Then ≡ k 1 1 − πℓ πhℓ s(h, k) = cot cot . 4k = k k Xℓ 1 32. Lecture 205

The reciprocity formula can be tackled immediately by the powerful method of residues. We have to construct the proper function for which these become the residues. Take πz πhz f (z) = cot πz cot cot k k and integrate over a rectangle with vertices iΩ, i(k+iΩ), indented at o and k. The poles of the ﬁrst factor all in the contour± one± 0, 1,..., k 1, for the second 0; and for the third 0, k/h, 2k/h,..., (h i)12/h. We have − − 1 ω2 cot ω = 1 ω − 3 −··· ! 278 About the triple pole z = 0,

1 k 1 π2z2 πz2 π2h2z2 f (z) = 1 + 1 + 1 + πz · πz · πhz − 3 ··· − 3 ··· − 3k2 ··· ! ! !

So the residue at z = 0 is

2 2 2 2 2 k π π π h = k k + 1 + h π2h − 3 − 3k2 − 3k2 −3π h hk k ! ! So

k 1 k k h 1 1 − πℓ πhℓ (Res) = + + + cot cot −3π h k hk π = k k X ! Xℓ 1 32. Lecture 206

h 1 k − πrh πh + cot cot πh = h h Xk 1 k k h 1 = + + + 12s(h, k) + 12s(k, h) 3π − h k hk ! ! And this is equal to 279 1 f (z)dz 2πi ZR where R is the rectangle. On the vertical lines the function the same value (by periodicity) end so the integrals cancel out. Hence

iΩ+k iΩ+k 1 1 − f (z)dz = 2πi R 2πi  −  Z  ZiΩ iZΩ   −  Now     iω + iω = e e− = + cot ω i iω iω , ω x iy, e e− − + eix y + e ix y = i − − ; eix y e ix+y − − − x varies from o to k and y = Ω, for this ± i, as y =Ω − →∞ uniformly →  i, as y = Ω  − → −∞ Therefore   1 1 lim f (z)dz = i3k ( i)3k Ω 2πi 2πi − − →∞ ZR 3n o = 2ki = k 2πi −π 280 k h k 1 ∴ + + + 12s(h, k) + 12s(k, h) 3π − k h hk ! ! = k −π k h 1 or 12s(h, k) + 12s(k, h) = 3 + + + , − h k hk ! which is the reciprocity formula. Part IV

Representation by squares

207 Lecture 33

We wish to begin the study of the representation of a number as the sum of 281 squares: n = n2 + n2 + + n2 1 2 ··· r We shall develop in this connection the Hardy-Littlewood circle method. Historically it is an off shoot of the Hardly-Ramanujan method in partition- theory, though we did not develop the latter in its original form in our treatment. The circle method has been applied to very many cases, and the problem of squares is a very instructive one for finding out the general thread. We shall later replace the problem by that of the representation of n by a positive quadratic form. This would involve only the general Poisson summation formula. In the case of representation as the sum of squares there is some simplification, because the generating ing function is the rth power of a simple V function. We shall deal with the asymptotic theory. Later we may go into Siegel’s theory of quadratic forms. Let us write ∞ 2 ∞ 2 Θ(x) = xn = 1 + 2 xn , n= n=1 X−∞ X x < 1. For r at least equal to 4, we consider | | r r ∞ n2 ∞ n2+n2+ +n2 Θ (x) = x = x 1 2 · r

n=  n j=  X−∞  X−∞   ∞  n = Ar(n)x , = Xn 0 on collecting the terms with exponent n, where Ar(n) is the number of times n 282

208 33. Lecture 209

appears as the sum of r square:

Ar(n) = 1 n2+ +n2=n 1 X··· r

It is clear that ni can be positive or negative. The more serious thing is that we have to count the representations diﬀerently when the summands are interchanged, in contradiction to the situation in the case of partitions. The problem of partition into squares would be a more complicated problem; the generating function would be more complicated, and what is worse, all the help one gets in partition theory from the theory of modular forms would break down here. th An(n) is the n coeﬃcient of a power-series; 1 Θr(x) A (n) = dx r 2πi xn+1 ZC where C is a suitable circle inside and close to the unit circle. The trick of 2πδ Hardy and Littlewood was to break the circle x = e− N where N is the order of a certain Farey dissection, into Farey arcs and| | write 1 Θn(x) A (n) = ′ dx, r 2πi xn+1 o h

2πi h 2πξ x = e k −

Rez < 0, and set z = δN iϕ, so chat we have a little freedom along both real and imaginary axes. − 2πi h 2πδ +2πiϕ x = e k − N . h The choice of the little arc ϕ is also classical. k is a certain Farey fraction, with adjacents h1 and h2 , say. h1 < h < h2 . We limit ϕ on the seperate arcs. k1 k2 k1 k k2 Introduce the mediants: h h + h h h + h h 1 < 1 < < 2 < 2 , k1 k1 + k k k2 + k k2 + + So that the interval h1 h , h2 h gives the movement of h + ϕ. So ϕ runs k1+k k2+2 k between + + V = h1 h h h2 h h = V hk′ ϕ hk′′ − k1 + k − k ≤ ≤ h2 + k − k 33. Lecture 210

V = 1 V = 1 hk′ ; hk′′ ; − −(k1 + k)k (k2 + k)k + and since 2N > k1 k > N, we have necessarily k2+k

1 V 1 2Nk ≤| hk|≤ Nk Now changing the variable of integration to ϕ, we can write

V hk′′ h h ′ 2πi n r 2πi 2πz 2πnz Ar(n) = e− k Θ e k − e dϕ 0 h

2πi h 2πz ∞ (2πi h 2πz)n2 Θ e k − = e k − · n= X−∞ k 1 − 2πi h j2 2πzn2 = e k e− j=0 n j (mod k) X ≡ X k 1 − 2πi h j2 ∞ 2πzk2(q+ j )2 = e k e k , j=0 q= X X−∞ where we have written n = kq + j. We can now handle this from our V -series formula. We proved (Lecture 12) that

C(τ) 1 V 2V V V (V / ) = eπiτ 3( τ/τ) i 3 − τ C(τ) i and = , Imτ> 0. i rτ Since 285 ∞ πin2τ 2πinV V3(V /τ) = e e , n= X−∞ writing τ = it, Ret > 0, we have from the above,

2 1 ∞ π n 2πinV πtV 2 ∞ πn2t 2πV nt e− t e = e− e− e− √t n= n= X−∞ X−∞ 33. Lecture 211

∞ πt(n+V )2 = e− n= X−∞ V j 2 Replacing n by q, by k and t by 2zk , we have

k 1 2 h − h 2 ∞ πq j 2πi 2πz 2πi j i 2 2πiq Θ e k − = e k e− 2zk e k 2 j=0 2zk q= X X−∞ ∞ πq2 1 p 2 = e− 2zk Tq(h, k) k √2z q= X−∞ k 1 2+ − 2πi hj qj where Tq(h, k) = e k = Xj 0

This is already a good reduction. Tq(h, k) depends on q modulo k, so it is periodic. We shall approximate to it in general. One special case, however, is of interest: for q = 0, 286

k 1 − 2πi h j2 T0(h, k) = e k = G(h, k), = Xj 0 where G(h, k) are the so-called Gaussian sums which we shall study in detail. They are sums of roots of unity raised to a square power, Θ is actually a V3, and when we evaluate Tq we get some other V . We now write

2πi h 2πz 1 Θ e k − = G(h, k) + H(h, k; z) k √2z { } ∞ πq2 2 where H(h, k; z) = Tq(h, k)e− 2k z q= Xq,−∞0

We shall throw H into the error term. Let us appraise Tq(h, k), not explicitly: that will take us into Gaussian sums.

k 1 2+ − 2πi hj qj Tq(h, k) = e k = Xj 0 k 1 k 1 2 − − 2πi (h j2+q j) 2π i (hℓ2+qℓ) Tq(h, k) = e k e− k | | = = Xj 0 Xℓ 0 33. Lecture 212

2π i (h(ℓ j2 ℓ2)+q( j ℓ)) = e k − − j Xmod k ℓ Xmod k 2π i ( j ℓ)(h( j+ℓ)+q) = e k − , j Xmod k ℓ Xmod k which, an rearranging according to the diﬀerence j ℓ, becomes 287 − 2πi a (h( j+ℓ)+q) = e k a mod k j ℓ a (mod k) X − ≡ X 2πi a (h(a+2ℓ)+q) = e k a Xmod k ℓ Xmod k 2π i (ha2+aq) 4πia h ℓ = e k e k a Xmod k ℓ Xmod k The inner sum is a sum of the roots of unity. Two cases arises, according as k 2a or k ∤ 2a. k odd implies that a = 0 and k even implies that a = 0 or k 2.| In case k 2a, the sum is zero. We then have | | i k2 + k 2 2π k h 4 2 q Tq(h, k) = k, if k is odd; k 1 + e , if k is even | | ! πi hk +q = k 1 + e ( 2 ) , if k is even = oor 2k if k is even = hk + It is of interest to notice that Tq 0 only if k is even and 2 q is an odd integer. In any case, T (h, k) √2k, | q |≤ and this cannot be improved. We then have

2 ∞ πq R 1 H(h, k; k; z) 2 √2ke− 2k2 z (q = 0 is not involved here). | |≤ = Xq 1 2 π 1 (q 1) 1 R ∞ π − R = 2 √2ke− 2k2 z e− 2k2 z = Xq 1 π R 1 ∞ 3πm R 1 = 2 √2ke− 2k2 z e− 2k2 z = Xm 0 π R 1 1 = 2 √2ke− 2k2 z 3π R 1 1 e− 2k2 z − Since z = δ iϕ, 288 N − 33. Lecture 213

1 R 1 = R 1 = R 1 = δN k2 z k2z k2(δ iϕ) k2(δ2 + ϕ2) N − N 1 1 δN 1 ∴ R , since Vhk , k2 z ≥ 2 2 + 1 | |≤ kN k δN N2 δN = 1 2 2 + 1 2 + 1 ≥ N δ 2 N δN 2 N N N δN We want to make this keep away from 0 as far as possible. This gives a + 1 desirable choice of δN . Make the denominator as small as possible. Since x x is minimised when x = 1, we have 1 1 1 R , k2 z ≥ 2

2 this minimum corresponding to N δN = 1. So if we choose the radius of the circle in terms of the Farey order, we shall have secured the best that we can:

π R 1 H(h, k; z) 2 √2ke− 2k2 z C | |≤ × It would be unwise to appraise the remaining exponential now. Lecture 34

2πi h 2πz We had discussed the sum Θ e k − and written it equal to 289

1 G(h, k) + H(h, k; z) k √2z { } π R 1 where H(h, k; z) < C √ke− 2k2 z | | If we apply this to the integral in which Θr appears,

r 2πi h 2πz r 1 r r λ Θ e k − = G(h, k) − H(h, k; z), r k λ k (2z) 2 λ=0 X ! or, keeping the piece corresponding to λ = 0 apart,

r h z r 1 1 r Θ 2πi k 2π r = r λ λ e − r G(h, k) r G(h, k) − H(h, k; z) − r z 2 r z 2 λ k (2 ) k (2 ) λ=1 X ! Let us appraise this. Since

r 2πi h 2πz r 1 r 1 r λ λ 2π 1 Θ k 2 2k2 e − r G(h, k) < C r ( √k) − k e− R − kr(2z) 2 kr z 2 z | | λ=1 X 1 π π R 1 2 2k2 z < C r e− e− · (k z ) 2 | | Now 290 V hk′′ 2πi h n 2πi h n 2πz Ar(n) = ′ e− k 2πznΘ e k − 0 h

V hk′′ 2π n r 2π n r C e N2 N 2 dϕ = Ce N2 N 2 , o h

V hk′′ e2πnz k z 4 VZ hk′ and write it as V hk′ ∞ ∞ − e2πnz dϕ  − −  zr/2 Z VZ′′ Z  −∞ hk −∞    The inﬁnite integrals are conditionally convergent if r > 0 (because the numerator is essentially trigonometric), and absolutely convergent for r > 2, 292 34. Lecture 216

so that we take r at least equal to 3. Then

∞ 2πnz ∞ e 2π n dϕ dϕ e N2 r/2 2 r z ≤ (δ + ϕ2) 2 VZ VZ N hk′′ hk′′

∞ 2π n dϕ N2 e r 2 + 2 4 ≤ (δN ϕ ) Z1 2kN

V − hk′ (Here and in the estimate of the other integral , we make use of the V V −∞ fact that the interval from hk′ to hk′′ is neither too longR nor too short. This argument arises also in Goldbach’s problem and Waring’s problem). The right side is equal to

∞ 2 ∞ r 2 2π n N dϕ 2π n r 2 dψ N2 = N2 N − e r e N − 2 r/4 (1 + N4ϕ2) 4 (1 + ψ ) Z1 ZN − 2kN 2k ∞ 2π n r 2 dψ < e N2 N − ψr/2 ZN 2k This appears crude but is nevertheless good since ϕ never comes near 0; 1 2 + 2 1 N/2k > 2 , and the ratio of ψ to 1 ψ is at least 3 and so we lose no essential order of magnitude. The last integral is equal to

r +1 2π n r 2 N − 2 Ce N2 N − , r 3, 2k ≥ 2π n r 1 r 1 = Ce N2 N 2 − k 2 −

V − hk′ A similar estimate holds for also. So, 293 −∞R r ∞ 2πnz 2πi h n G(h, k) e k ϕ Ar(n) ′ e− r/2 d − 0 h

2π n r/2 2π h r/2 1 < Ce N2 N + Ce N2 N − ′ 0 k N ≤ ≤ 2π n r/2 = Ce N2 N . P This, however, does notgo to zero as N ; wehavenogoodluckhereas we had in partitions. So we make the best of→∞ it, and obtainan asymptotic result. Let n also tend to inﬁnity. We shall keep n/N2 bounded, without lotting; it go to zero, as in the latter case the exponential factor would become 1. We have to see to it that n CN2 i.e., N is at least √n. Otherwise the error term would ≤ increase fast. Making N bigger would not help in the ﬁrst factor and would 294 make the second worse. So the optical choice for N would be N = [ √N]. The error would now be r O n 4 We next evaluate the integral

∞ e2πnz dϕ zr/2 Z −∞ This is the some as

+ ∞ e2πn(δN iϕ) −∞ e2πn(δN iα) − dϕ = dϕ (δ iϕ)r/2 − (δ + iα)r/2 Z N − Z N −∞ ∞ δN +i 1 ∞e2πns = ds i sr/2 δ Zi N − ∞ After a little embellishment this becomes a well-known integral. It is equal to 2πnδN +i (2πn)r/2 ∞ eω dω i ωr/2 2πnδZ i N − ∞ which exists for r > 2, and is actually the Hankel loop integral, and hence equal to 2π(2πn)r/2 1 − Γ(r/2) Hence, for f 3. We hence the number of representations of n as the sum 295 of r squares: ≥

r/2 r 1 r (2π) n 2 − G(h, k) 2ri h r/4 = k + . Ar(n) r/2 ′ r e− O(n ) Γ(r/2) · 2 · 0 h

One ﬁnal step. Let us improve this a little further. Write

r G(h, k) 2πi h n (r) e− k = V (n) = V (n) kr k k h Xmod k

We have to sum Vk(n) from k = 1 to k = N. However, we sum from k = 1 to k = , thereby incurring an error ∞

∞ ∞ r +1 Vk(n) k− 2 , = + ≤ = + kXN 1 kXN 1 and this converging absolutely for r 5 is ≥ r +2 r +1 O N− 2 = O n− 4

r 1 r/4 This along with the factor n 2 − would give exactly O(n ). (We could have saved this for r = 4 also if we had been a little more careful). Thus, for r 5, ≥ we have 296

r/2 π r 1 r/4 A (n) = n 2 − S (n) + O(n ), r Γ(r/2) r

∞ where S r(n) = Vk(n) = Xk 1 S r(n) is the singular series. We shall show that S r(n) remains bounded at least for r 5. ≥ Lecture 35

After we reduced our problem to the singular series in which the Gaussian 297 sums appear conspicuously, we have to do something about them before we proceed further. The Gaussian sums are deﬁned as

2πi h ℓ2 G(h, k) = e k , (h, k) = 1 ℓ Xmod k

They obey a simple multiplication rule: if k = k1k2,(k1, k2) = 1, then

G(h, k k ) = G(hk , k ) G(hk , k ). 1 2 1 2 · 2 1

For, put ℓ = rk1 + sk2; when r runs modulo k2 and s modulo k1,ℓ runs through a full residue system modulo k1k2. Hence

h + 2 2πi k k (k1r k2 s) G(h, k1k2) = e 1 2 k Xmod k2 s Xmod k1 2πi h (k2r2+k2 s2) = e k1k2 1 2 r Xmod k2 s Xmod k1 2πi hk1 r2 2πi hk2 s2 = e k2 e k1 r Xmod k2 s Xmod k1 = G(hk1, k2)G(hk2, k1).

Ultimately, therefore, only prime powers have to be considered to denominators. We have to distinguish the cases p = 2 and p > 2, p prime. 1) Let p 3, k = pα with α> 1 298 ≥ α 2πi h ℓ2 G(h, p ) = e pα r l Xmod p 219 35. Lecture 220

α 1 write ℓ = mp − + r;

m = 0, 1,..., p 1; r = 0, 1,... pα 1 1. Then this becomes − − − α 1 α 1 1 p 1 p − 1 p 1 p − − − − 2πi h (mpα 1+r)2 − 2πi h (m2 p2α 1+2mrpα 1+r2) e pα − = e pα − − = = = = Xm 0 Xr 0 Xm 0 Xr 0 Since α 2, 2α 2 α and so the ﬁrst term in the exponent may be omitted. This≥ gives − ≥ α 1 p − 1 p 1 − 2πi h r2 − 2πi h 2mr e pα e p = = Xr 0 Xm 0 The inner sum is a sum of pth roots of unity; so it depends on whether p divides 2rh or not. But (h, p) = 1 and p ∤ 2. So we need consider only the cases: p r and p ∤ r. However in the latter case this sum is 0 while in the former it is| p. We therefore get, when p r, r = ps, | α 1 α 2 p − 1 p − 1 − 2πi h r2 − 2πi h p2 s p e pα = p e pα r=0,p r s=0 X| X α 2 p − 1 − h 2πi α 2 s = p e p − s=0 X α 2 = pG(h, p − )

We have therefore reduced the never of the denominator by 2. We can 299 repeat the process and proceed as long as we end with either the0th or the 1st power. So we have two chances. In the former case, evidently G(h, 1) = 1. So for α even, G(h, pα) = pα/2 On the other hand, if α is odd, we have

α α 1 G(h, p ) = p −2 G(h, p).

These may be combined into the single formula

α [ α ] α 2[ α ] G(h, p ) = p 2 G h, p − 2 (1) 2) p = 2λ, λ 2. h is now odd. Write ≥ λ 1 λ 1 ℓ = m2 − + r; m = 0, 1; r = 0, 1,..., 2 − 1 − 35. Lecture 221

λ 1 λ 1 2 − 1 2 − 1 λ − 2πi h r2 − 2πi h (2λ 1+r)2 G(h, 2 ) = e 2λ + e 2λ − = = Xr 0 Xr 0 since λ 2, 2λ 2 λ, in the second sum it is only the exponent r2 that ≥ − ≥ contributes a non-zero term; and this is then the same the ﬁrst. Altogether we 300 have then λ 1 2 − 1 − πi h r2 2λ 1 2 e − (*) = Xr 0 This, however is not a Gaussian sum. The substitution for ℓ does not work; to beeﬀective, then we take

λ 2 λ 2 ℓ = m2 − + r; m = 0, 1, 2, 3; r = 0, 1,..., 2 − 1. − Now take λ 4 and start again all over. ≥ λ 2 3 2 − 1 λ − 2πi h (m2λ 2+r)2 G(h, 2 ) = e 2λ − = = Xm 0 Xr 0 λ 2 3 2 − 1 − 2πi h (2λ 1mr+r2) = e 2λ − , (for λ 4 i.e., 2λ 4 λ). = = ≥ − ≥ Xm 0 Xr 0 λ 2 2 − 1 3 − 2πi h r2 πihmr = e 2λ e = = Xr 0 Xm 0 λ 2 2 − 1 3 − 2πi h r2 mn = e 2λ ( ) = = − Xr 0 Xm 0 λ 2 λ 2 2 − 1 2 − 1 − r 2πi h r2 − 2πi h r2 = 2 ( ) e 2λ + 2 e 2λ = − = Xr 0 Xr 0 λ 3 2 − 1 − h 2 πi λ 3 s = 4 e 2 − = Xs 0 This is not Gaussian sum either. But is is of the form (*). We therefore 301 λ λ 2 have, for λ 4, G(h, 2 ) = 2G(h, 2 − ). If λ = 4, we need go down to only 22 = 4 and if≥λ = 5to23 = 8. So we need separately G(h, 8) and G(h, 4); and of course G(h, 2). These cases escape us, while formerly only G(h, p) did. For λ 4, we may write ≥ λ [ λ ] 1 λ 2[ λ +2] G(h, 2 ) = 2 2 − G h, 2 − 2 (2) 35. Lecture 222

This supplements formula (1). We now consider the special cases, k = 2, 4, 8. Here h is odd.

2πi h G(h, 2) = 1 + e 2 = 0 2πi h 1 2πi h 4 2πi h 9 G(h, 4) = 1 + e 4 · + e 4 · + e 4 · πi h = 2 + 2e 2 = 2 1 + ih

πih 2πi h G(h, 8) = 1 + 1 + 2e + 4e 8 (since 12, 32, 52, 72 are all 1 modulo 8) ≡ + 2 πi h 1 i = 4e 4 = 4 √2 ! Before we return to G(h, p), p > 2, we shall a digression an connect to the 302 whole thing with the Legendre-Jacobi symbols

p 1 − 2πi h ℓ2 G(h, p) = e p = Xℓ 0 2πi h a = 1 + 2 e p , a X the summation over all quadratic residues a modulo p, since along with ℓ, p ℓ is also a quadratic residue. We can write this in a compact form, so arranging− it that the non-residues get cancelled and the residues appear twice:

r 2πi h r G(h, p) = 1 + e p p r Xmod p ( !) r 2πi h r = e p p r Xmod p ! This would appear in a completely new aspect if we utilised the fact that hr runs through a full system of residues modulo p. Then

h hr 2πi h r G(h, p) = e p p p k Xmod p ! ! h r 2πi r = e p p p ! r Xmod p ! 35. Lecture 223

h = G(h, p). p ! This is very useful if we new go to the Jacobi symbol. For prime p, the 303 Legendre symbol has the multiplicative property:

r1 r2 = r1r2 p p p ! ! ! Jacobi has the following generalisation. r Deﬁne pq by r r r = . pq p q ! ! ! Si it is 1; if it is +1 it does not necessarily mean that r is a quadratic residue modulo pq±. The Jacobi symbol no longer discriminates between residues and non residues. From the deﬁnition then

a a α a β = . pαqβ p q ··· ···! ! ! The Jacobi symbol has the properties of a character, as can be veriﬁed by using the Chinese remainder theorem. We can now write h α G(h, pα) = G(1, pα) p ! under all circumstances. How does this come about? Separate the cases: α even, α odd.

G(g, pα) = G(1, pα), α even; α 1 = p 2− G(h, p), α odd,

h α 1 h α = p −2 G(1, p) = G(1, p ) p p ! ! We can write both in one sweep as 304

h α G(h, pα) = G(1, pα) p ! h = G(1, pα) pα ! 35. Lecture 224

Now use the multiplicative law. If p, q are odd primes, then

G(h, pαqβ) = G(hpα, qβ)G(hqβ, pα) hpα hqβ = G(1, qβ) G(1, pα) qβ pα ! ! Since the Jacobi symbol is separately multiplicative in numerator and denominator, but not both, this is equal to

h pα h qβ h h G(1, qβ) G(1, pα) = G(pα, qβ)G(qβ, pα), qβ qβ pα pα qβ pα ! ! ! ! ! ! taking the second and third factors together, and also the last two. And this is

h G(1, pαqβ) pαqβ ! according to the multiplication law. Suppose that we have

h h G(h k ) = G(1, k ); G(h, k ) = G(1, k ). 1 1 k 1 2 k 2 1 ! 2 ! We go through the above worker; literally and get 305

h G(h, k k ) = G(1, k , k ). 1 2 1, h 1 2 2 ! So we have proved in general that for odd k,

h G(h, k) = G(1, k) k ! We can now return to G(h, p). Lecture 36

We were discussing Gaussian sums and it remained to evaluate 306

h G(h, p) = G(1, p) p ! We shall do a little more than that; we shall study them in a more ﬂexible form. Deﬁne k 1 − πi hg ℓ2 S (h, k) = e k , = Xℓ 0 h, k > 0 but not necessarily coprime. We cannot now take the summation over ℓ modulo k. For if k is odd, (ℓ + k)2 = ℓ2 + 2ℓk + k2 and k2 may give rise to an odd multiple of πi in the exponent and hence introduce a change of sign, We should therefore insist on this particular range of summation. S (h, k) are connected with the Gaussian sums; indeed

G(h, k) = S (2h, k)

We shall now produce S (h, k) as a sum of residues. To get the integers as poles we should clearly take e2πiz 1 in the denominator; so we integrate πi h z2/(e2πiz 1) − e k − over such a contour as has in its interior the desired poles z = 0, 1, 2,..., k 1. Indeed − πi h z2 e k S (h, k) = dz e2πiz 1 ZC −

225 36. Lecture 226

0 1 2

Where C is the parallelogram with vertices at (1 + i)Ω, (1 + i)Ω+ k, 307 ± ± with the slant sides inclined at 45◦ (infact this may be anything less than 90◦) to the real axis, and making a detour round 0 and k. When we push Ω to , the integrals along the horizontal sides will tend to zero. For instance on the∞ upper side, z = (1 + i)Ω+ x,0 x k, and the integrand is therefore ≤ ≤ πi h ((1+i)Ω+x)2 πi h (2iΩ2+2(1+i)Ωx+x2) e k = e k 2πi((1+i)Ω+x) 2πi(Ω+x) 2πΩ e 1 e − 1 − π h (2Ω2+2Ωx)+π−i h (2Ωx+x2) = e− k k e 2πΩ+2πi(Ω+x) 1 − − 0 uniformly as Ω since h > 0. Hence the integral can be written as 308 → →∞ k (1+i) +k (1+i) ∞ ∞ πi h (z)2 e k dz − e2πz 1 (1+Zi) +k (1Z+i) − − ∞ − ∞ where, of course, we have to make a small detour round 0 and k. Replacing z by z + k in the ﬁrst integral, this becomes

(1+i) (1+i) h 2 h 2 h 2 h 2 z z + ∞ πi (z+k) πi z ∞ πi k πi k (2 k k ) e k e k e e 1 − dz = − dz e2πiz 1 e2πiz 1 (1Z+i) − (1Z+i) − − ∞ − ∞ (1+i) ∞ π h z2 e2πihz+πihk 1 e k ( ) = − dz e2πiz 1 (1Z+i) − − ∞ 36. Lecture 227

Let us assume from now on that hk is even. Then we can actually divide 309 out and the integral becomes

(1+i) ∞ h 1 πi h z2 − 2πiλz e k e dz  =  (1Z+i) λ 0 − ∞  X    The denominator has now disappeared. There is a further advantage that the integral can now be stretched along the whole line and the detour can be avoided. We then have

(1+i) h 1 ∞ 2 − πiλ2 h πi h (z+ λk ) e− k e k h dz = λ 0 (1Z+i) X − ∞ Write z + λk/h = ω; and shift the integral back to the line from (1 + i) to (1 + i) - this we can do since the integrand tends to zero along a− horizontal∞ segment.∞ This gives (1+i) h 1 ∞ − πi h λ2 πi h ω2 e− k e k dω, = λ 0 (1Z+i) X − ∞ = h h or writing t ω k , k > 0, 310 q q (1+i) h 1 ∞ h 1 k − πi h λ2 πit2 k − πi k λ2 e− k e dt = A e− h h = h = r λ 0 (1Z+i) r λ 0 X − ∞ X where A is the speciﬁc constant:

(1+i) ∞ 2 A = eπit dt

(1Z+i) − ∞ Hence k S (h, k) = A S ( k, h). rh − In order to evaluate A, take a simple case: h = 1, k = 2

S (1, 2) = A √2S ( 2, 1) πi − i.e., 1 + e 2 = A √2, 36. Lecture 228

So A = (1 + i)/ √2, an eighth root or unity. So our reciprocity formula becomes complete:

1 + i k S (h, k) = S ( k, h). √2 rh − Let us develop some corollaries. 1) h = 2, k arbitrary: 311

S (2, k) = G(1, k), so 1 + i k G(1, k) = S (2, k) = S ( k, 2) √2 r2 − + 1 i πi k = √k(1 + e− 2 ) 2 1 + i = √k(1 + ( i)k) 2 − We then have explicitly the value of G(1, k)

(1 + i)(1 + ( i)k) G(1, k) = − √k. 2 We mention the four cases separately:

√k if k 1 (mod4) ≡  0 if k 2 (mod4) G(1, k) =  ≡  i √k if k 3 (mod4)  ≡  + (1 i) √k if k 0 (mod4)  ≡  Hence the absolute value ofG(1, k)canbe0, k or √2k. So far k was only positive. The case k odd deserves some special mention. k 1 is even and − √k if k 1 is even G(1, k) = −2  k 1 i √k if − is odd.  2 2  k 1  k 1 −2 0, 1 (mod 4) accordingas −2 is even or odd; so we can write this 312 as ≡ k 1 2 G(1, k) = i( −2 ) √k. 36. Lecture 229

This we have obtained by a purely function-theoreticalargument. From our arithmetical augment, we had, for odd k,

h k 1 2 G(h, k) = i( −2 ) √k k ! h where k is the Jacobi symbol. We can get a little more out of it. 1 k 1 2 G( 1, k) = − i( −2 ) √k. − k ! Multiplying this and the equation for G(1, k) together,

1 k 1 2 G(1, k)G( 1, k) = − ( )( −2 ) k − k − ! 1 k 1 = − ( ) −2 k k − ! But the left side is only G(l, k)G(1, k), and this is always > 0. So

1 k 1 − ( ) −2 k > 0, k − ! and since k > 0 by nature, 1 k 1 − = ( ) −2 k − ! which is Euler’s criterion for the Jacobi symbol. 2) h = 2, k odd.

1 + 1 k G(2, k) = S (4, k) = b S ( k, 4) √2 r4 − + 1 i πik πik πik = √k 1 + e− 4 + e− + e− 4 2 √2 + n o 1 i 1 πi k = √2 √ke− 4 √2 i πi (k 1) = e− 4 − √k

πi k 1 − √2 = e− 2 2 k 1 − = i− 2 √k 36. Lecture 230

On the other hand 313 2 k 1 2 G(2, k) = i( −2 ) √k k ! Hence

2 k 1 k 1 2 − ( − ) = i− 2 − 2 k ! k 1 k 1 − (1+ − ) = i− 2 2

k2 1 − = i− 4

k2 1 2 − = i− 8

k2 1 = ( ) 8− − 3) (h, k) = 1; h, k both odd:

1 + i k G(h, k) = S (2h, k) = S ( k, 2h) √2 r2h − + 2h 1 1 i k − πi k λ2 = e 2h 2h √2 r λ=0 X + 1 i k πi k λ2 = e− 2h 2h √2 r λ mod 2h X Here it is no longer necessary to insist on the special range of summation, 314 for changing λ by λ + 2h would introduce only an even multiple of πi in the exponent. Separating the odd and even λ′ s, this becomes + 1 i k πi k (2ℓ)2 πi k (2ℓ+h)2 e− 2h + e− 2h 2h √2 r ℓ mod h ℓ mod h   X X   +   1 i k πi hk 2πi k ℓ2 = 1 + e− 2 e− h  2h √2 r ℓ mod h X 1 + i k = 1 + ( i)hk G( k, h) √2 − r2h − hk 1 2 k = i( 2− ) G( k, h) rh − hk 1 2 k = i( 2− ) G(k, h) rh 36. Lecture 231

Then we have 315

h k 1 2 hk 1 2 k h h 1 2 ( − ) ( − ) ( − ) i 2 √k = i 2 i− 2 √h k h k ! r ! h k hk 1 2 h 1 2 k 1 2 ( − ) ( − ) ( − ) i.e., = i 2 − 2 − 2 k h ! ! ib 1 where b = h2k2 h2 k2 + 1 2(hk h k 1) 4 − − − − − − 1 = (h 1)(k 1) (h + 1)(k + 1) 2 4 − − { − } 1 (h + 1)(k + 1) = [(h 1)(k 1)] 1 2 − − 2 − " # So

(h 1)(k 1) b 2 − − i = i 4 an odd number (h 1)(k 1) (h 1)(k 1) = ( ) − 4 − (odd number) = ( ) − 4 − − − h k (h 1)(k 1) ∴ = ( ) − 4 − . k h − ! ! which is Jacobi’s law of reciprocity. We shall use all this in the singular series. It may be worth while to do what 316 Gauss himself did and evaluate G(1, k) by an arithmetical method. To distinguish between the diﬀerent primitive roots of unity is, however, algebraically impossible; in the analytical method we can use the exponential function to uniformise the roots of unity. Lecture 37

We have ﬁnished to some extent Gaussian sums; we treated then only in view 317 of their occurrence in the singular series deﬁned as

(r) = ∞ (r) S (n) Vk (n) k=1 X r (r) G(h, k) 2πi h n with V (n) = V (n) = e− k , k k k h mod k ! (hX,k)=1 which appeared as the principal term in the expression for the number of representation of n as the sum of r squares:

r/2 π r 1 (r) r/4 Ar(n) = n 2 − S + O n , Γ r (n) 2 r 5. We did not bother to do this for lower r, although we could for r = 4, in≥ which case we know an exact formula; but this is another question. We consider ﬁrst a fundamental property of the singular series, viz. its expression as an inﬁnite product. Fundamental Lemma.

(r) = + + + + S (n) 1 Vp(n) Vp2 (n) Vp3 (n) , p ··· Y n o p prime. We ﬁrst prove the multiplicative property of Vk(n): for (k1, k2) = 1, = Vk1 (n)Vk2 (n) Vk1k2 (n)

232 37. Lecture 233

We had a similar situation in connection with Ak(n) for the partition func- 318 tion; but there the multiplication was more complicated. Here we have

1 r 2πi hn = − k1k2 Vk1k2 (n) r G(h, k1k2) e . (k1, k2) h mod k1k2 (h,kX1k2)=1

Writing h = k2h1 + k1h2 with the conditions (h1, k1) = 1 = (h2, k2), h, running modulo h1 and h2 modulo k2, this becomes

1 2πi h n + − k1k2 r G(k2h1 k1h2, k1k2)e (k1k2) Xh1 Xh2 = 1 + r r G ((k2h1 k1h2)k1, k2) (k1k2) h1 h2 X X + n r 2πi(k2h1 k1h2) k k G((k2h1 + k1h2)k2k1) e− 1 2 on using the multiplicativity of the Gaussian sums; and suppressing multiples of k1, k2, as we may, this gives

h h 1 2 r 2 r 2πi 1 n 2πi 2 n k1 k2 r r G(h2k1, k2) G(k2h1, k1) e− − k1k2 h1 Xmod k1 h2 Xmod k2 Now 319 2 2πi h a2ℓ2 G(ha , h) = e k ℓ Xmod k If (a, k) = 1, al also runs modulo k when ℓ does, so that the right side is

2πi h m2 e k = G(h, k) n Xmod k In our case (k1, k2) = 1. So we have

h h 1 r 2πi 1 n 1 r 2πi 2 n k1 k2 r G(h1, k1) e− r G(h2, k2) e− k1 k2 h1 Xmod k1 h2 Xmod k2 = Vk1 (n)Vk2 (n)

(r) We can then break each summand in S n into factors corresponding to prime powers and multiply them again together, and the rearrangement does not count because of absolute convergence; so (r) = + + + + S (n) 1 Vp(n) Vp2 (n) Vp3 (n) p ··· Y n o 37. Lecture 234

= γp(n), p Y say; this is an absolutely convergent product. This simpliﬁes matters consid- 320 erably. We have to investigate V only for those G′ s is which prime powers appear. We ﬁrst take p = 2. then

γ (n) = 1 + V (n) + V 2 (n) + 2 2 2 ··· 1 λ r 2πih n V λ (n) = G(h, 2 ) e− 2λ 2 2λr h mod 2λ X2∤h (i) λ = 1 Since G(h, 2) = 0 for odd h,

V2(n) = 0 (ii) λ even. For λ 4, ≥

λ λ 1 h λ h G(h, 2 ) = 2 2 − 2(1 + i ) = 2 2 (1 + i )

1 λr/2 h r 2πi h n V λ (n) = 2 (1 + i ) e− 2λ 2 2λr h mod 2λ X2∤h

1 r 2πi h n r 2πi h n = (1 + i) e− 2λ + (1 i) e− 2λ 2λr/2  −  h 1 (mod 4) h (mod 4)   ≡h Xmod 2λ ≡−h Xmod 2λ    r/2   2  πi r 2πi h n πi r 2πi h n  = e 4 e− 2λ + e− 4 e− 2λ 2λr/2   h 1 (mod 4) h 1 (mod 4)   ≡ X ≡− X  1  πi r 2πi r 2πi s n  =  4 2λ 2λ 2 +  λ 1 e − e− −  − r 2 2 λ 2  s mod 2 −  X  r + r s  πi 4 2πi λ 2πi λ 2 n  +e− 2 e− 2 − λ 2  s modX 2 −  λ 2  = 0, if 2 − + n;  λ 2 2 − r ν λ 2 λ 2 cos π 2π , if 2 − /n, n = 2 − .ν λ 1 r 2 −2 4 − 4 1 π i.e., cos (2ν r) (λ 1)( r 1) 2 − 4 − 4 − 37. Lecture 235

321 Hence, for λ even, λ 4, ≥

λ 2 0, if 2 − + n; V2λ (n) = (*)  cos π (2ν r) 4 − λ 2 =  (λ 1)( r 1) , if 2 − .ν n.  2  2 − −  (iii) λ odd, λ 3.  ≥ λ 3 λ λ 2 − 3 G(h, 2 ) = 2G(h, 2 − ) = 2 2 G(h, 2 ) λ 3 πih/4 λ+1 πih/4 = 2 −2 4e = 2 2 e

1 λ+1 πih r 2πi h n V λ (n) = 2 2 e 4 e− 2λ , 2 2λr h mod 2λ X2∤h or, writing h = 8s + t, t = 1, 3, 5, 7, 322

λ 3 2 − 1 πitr/4 2πi(8s+t) n = 2λ λ 1 e e− − r 2 2 t = X Xs 1 λ 3 2 − 1 λ 3 = πitr/4 2πitn/2λ 2πisn/2 − λ 1 e − e− − r 2 2 t s=1 Xλ 3 X = 0, if 2 − ∤ n.

If, however, 2λ 3 n, n = 2λ 3.ν, this is − | − λ 3 2 − πit∤4(r ν) = λ 1 e − o, if 4/(r ν); − r 2 2 t − λ 1 X 2 − πi(r ν) 4 e − | , if 4/(r ν) λ 1 r 2 −2 − 1 ν r i.e., ( ) −4 . (λ 1)( r 1) 2 − 2 − · − Hence for λ odd, λ 3, 323 ≥

λ 3 0, if 2 − ∤ n;  λ = λ 3 λ 3 V2 (n)  0, if 2 − . n, n = 2 − ν, 4 ∤ (ν r); (**)  | −  ν r  ( ) −4 λ 3 (λ−1)( r 1) , if 2 − n, 4 (ν r)  2 − 2 − | | −    37. Lecture 236

Now, given n, only a ﬁnite number of powers of 2 can divide it. So the λ 3 situation 2 − /n will occur sometime or the other, so that γ2(n) is always a ﬁnite sum.

∞ 1 γ (n) 1 2 (λ 1)( r 1) | − |≤ = 2 − 2 − Xλ 2 = 1 1 r 1 r 2 2 · 1 1/2 1 − − 2 − 1 = ; 2r/2 1 1 − − and this is valid for r 3. so the singular series behaves much better than we expected. ≥ Lecture 38

It would be of interest to study γ2(n) also for r = 3, 4. 324

γ (n) = 1 + V (n) + V 2 (n) + 2 2 2 ··· First consider the case r = 3, 2/n. Then

V2(n) = 0.

For V2λ (n), λ> 1, we have to make a distinction between λ even and λ odd. λ even.

λ 2 0, if 2 − ∤ n; V2λ (n) =  cos π (2ν r) 4 − λ 2 = λ 2  (λ 1)( r 1) , if 2 − ∤ n, n 2 − .ν.  2  2 − − λ odd.  

λ 3 0, if 2 − ∤ n;  λ = λ 3 λ 3 V2 (n)  0, if 2 − n, n = 2 − ν, ν r . 0 (mod4)  | −  ν r  ( ) −4 λ 3 = λ 3 (λ−1)( r 1) , if 2 − n, n 2 − ν, ν r 0 (mod4)  2 − 2 − | − ≡   So for r = 3, 

γ2(n) = 1 + V4(n) + V8(n) π n 3 cos (2n 3) ( ) −4 = 1 + 4 − + − , √2 2

237 38. Lecture 238

where the last summand has to be replaced by 0 if (n 3)/4 is not an integer. 325 Since 2n 3 is odd, we have − − π 1 cos (2n 3) = , | 4 − | √2 and thus clearly, 1 1 γ (n) 1 + + = 2 | 2 |≤ 2 2 Moreover, γ2(n) can vanish. This would require

n 3 ( ) −4 = 1 − π 1 and cos (2n 3) = 4 − − √2 simultaneously. But this is the case for

n 7 (mod8), ≡ as is easily seen. This corresponds to the fact that a number n, n 7 (mod 8) cannot be represented as the sum of three squares. ≡ Next take r = 4. We distinguish between the cases 2 ∤ n and 2 n. | 1. 2 ∤ n. Then from relations (*) and (**) proved in lecture 37, we have

γ2(n) = 1 + V4(n) + V8(n) cos π (2n 4) 1 πn = 1 + 4 − = 1 cos 2 − 2 2 = 1

α 2. 2 n Let n = 2 n , 2 n . Then (*) and (**) show that V λ (n) = 0 for 326 | ′ | ′ 2 λ > α + 3. But actually V2λ (n) = 0 also for λ = α + 3. Indeed, for α odd, λ = α + 1 is the last even, λ = α + 2 the last odd index for non-vanishing V2λ (n). For α even, λ = α + 2 is the last even index: λ = α + 3 is odd and since 4 ∤ (n 4), we have also V λ (n) = 0 for λ = α + 3. ′ − 2 α+2 α ∴ γ2(2 n′) = 1 + V2λ (n) = Xλ 2 Now, in V2λ (n), for λ even,

π π α λ+2 cos (2ν r) = cos n′2 − 4 − − 2 38. Lecture 239

α λ+1 = cos πn′2 − − 1, for λ α, − ≤ =  = + 1, for λ α 1,   = + 0, for λ α 2.   Similarly in V2λ (n), for λ odd, 

ν 4 n1.2α λ+1 ( ) −4 = ( ) − − − − 1, for λ α; = − ≤  1, for λ = α + 1,    α λ+1 and V2λ (n) = 0 for λ = α + 2 since then 4 ∤ 2 − . The numerators of the 327 non-vanishing V λ (n) are 1 upto the last one, which is 1. And thus 2 − 1 1 1 1 γ α = + 2(2 n′) 1 2 α 1 α − 2 − 2 −···− 2 − 2 = 1 + 1 = 3 α 1 α α 2 − 2 2 α Although here γ2(2 n′) > 0, we see that for α suﬃciently large γ2(n) can come arbitrarily close to 0. We now consider γ (n) for p 3. p ≥

γ (n) = 1 + V (n) + V 2 (n) + , p p p ··· h 1 λ r 2πi λ n where V λ (n) = G(h, p ) e− p p pλr h mod pλ Xp∤h Now h G(h, pλ) = G(1, pλ) pλ ! 2 λ pλ 1 h − λ = i 2 p 2 p ! 2 pλ 1 r − 2 h i h 2πi λ n ∴ V λ (n) = e− p p pλr/2 p h mod pλ ! Xp∤h 38. Lecture 240

We have to distinguish between λr odd; and λr even 328 1) λr even. If pλ 1 (mod 4), then ≡ λ 2 r p 1 λ − r p 1 ( ) 2 2 = ( ) 2 2− − − So 2 pλ 1 r − 2 h i 2πi λ n V λ (n) = e− p p pλr/2 h mod pλ Xp∤h 2) λr odd. In this case

2 pλ 1 r − 2 h i h 2πi λ n V λ (n) = e− p p pλr/2 p h mod pλ ! Xp∤h

The inner sum here is a special case of the so-called Ramanujan sums:

2πi h n Ck(n) = e k h mod k (hX,k)=1 There sums can be evaluated. Look at the simpler sums

2πi λ n S k(n) = e k λ Xmod k k, if k n; = |  0, if k ∤ n.   Classify the λ′ s in S k(n) according to their common divisor with k. Then 329

2πi λ n S k(n) = e k d k λ mod k X| (λ,Xk)=d 2πi λ n = e d · k/d d k λ mod k | Xλ kX= ( k , d ) 1 2πi µn = e k/d d k µ mod k X| X d k = (µ, d ) 1 38. Lecture 241

= C k (n) d d k X| = Cd(n). d k X| Now by M¨obious inversion formula,

k C (n) = µ S (n), k d d d k X| ! and S d(n) is completely known- it is either 0 or d; hence

k C (n) = dµ k d d k,d n X| | ! k = dµ . d d (n,k) X| ! So these are integers. 330 The M¨obious function which appears here arises as a coeﬃcient in a certain Dirichlet series; in fact 1 = ∞ µ(n) s ζ(s) = n Xn 1 It is possible to build up a complete formal theory of Dirichlet series as we had in the case of power series. Formal Dirichlet series form a ring without null-divisors. The multiplication law is given by

an bn = cn n2 n2 ns X X X where cn = akb j = kX j n The relation µ(n) 1 = 1 ns ns thenimpliesthatX 0 = Xµ( j) 1 = µ(d), n > 1. · jk=n d n X X| Lecture 39

For p 3 we had 331 ≥ γ (n) = 1 + V (n) + V 2 (n) + p p p ··· where

h 1 λ r 2π λ n V λ (n) = G(h, p ) e− p p pλr h mod pλ (hX,p)=1

2πi h n − pλ , λ λ 2r e r even; p 1 λ 2− h mod p i  ∤ =  pPh λr/2  p  2πi h n  h − pλ  p e , λr odd. h mod pλ  p∤h  P  For λr odd we have to evaluate this directly. If λr is even it is simpler; it is a special case of the Ramanujan sums:

2πi h n Ck(n) = e k h mod k (hX,k)=1 which could be evaluated by means of the M¨obious inversion formula: k C (n) = dµ k d d (k,n) X| ! So if k is a prime power, k = pλ, 332 λ 2πi h n p e− pλ = dµ d h mod pλ d (pλ,n) ! Xp∤h |X

242 39. Lecture 243

0, if α < λ 1, n = pαn , p ∤ n ; − ′ ′  λ 1 α  1 p − = p . if α = λ 1; = − × −− −   λ 1 λ  1 p − + p − ×  = pλ(1 1 ), if α λ.  − p ≥  For obtaining these values we observe that in the summation on the right  pλ side we have to take into account only such divisors d that α is at most p. This leads in the ﬁrst case α < λ 1 to a vacuous sum. In the second case the only λ 1 − λ 1 λ admissible divisor is p − ; in the last we have two divisors p − and p . Thus

Vpλ (n) = 0

for λ > α + 1; we get again a ﬁnite sum for γp(n) We now take λr odd. We want

h 2πi h n e− pλ p h mod pλ ! Xp∤h h modulo p is periodic, and we emphasize this by writing λ 1 h = rp + s; s = 1, 2,..., p 1; r = 1,..., p − − 333 So the above sum becomes λ λ p 1 p 1 + p 1 p 1 − − s 2πi (rp s) − s 2πi s − 2πi r n pλ = pλ pλ 1 e− e− e− − = = p = p = Xr 1 Xs 1 ! Xs 1 ! Xr 1 λ 1 This is zero when p − ∤ n (because the inner sum vanishes). Otherwise, λ 1 let n = p − ν and p ∤ ν; then it is again zero because we have only a sum of quadratic residue symbols (since the character is not the principal character). If p ν, the sum becomes | p 1 2 λ 1 λ 1 ν − p − G(ν, p) = p − i 2 √p p ! So if n = pα n where p ∤ n , then · ′ ′ 0, if λ 1 > α; −  p 1 2 = − Vpλ (n) pα n′ i 2 p, if λ 1 = α;  p √  −   0, if 0 λ 1 < α.  ≤ −   39. Lecture 244

So the only non vanishing term in the case α + 1 odd is Vpα+1 (n). Let us put things together now. Let r be even. If p ∤ n, then

p 1 2 − r i 2 γ = 1 + V = 1 p p − pr/2 r p 1 ( ) 2 −2 = 1 − − pr/2 334 If p n, n = pα n , then | · ′

γ = 1 + V + V 2 + + V α + V α+ p p p ··· p p 1 2 ǫp ǫp = 1 + (p 1) + p(p 1) + pr/2 − p2r/2 − ··· ǫα ǫα+1 p α 1 p α + p − (p 1) p , pαr/2 − − p(α+1)r/2 where ǫ = ( )r(p 1)/4 for r , 4 p − − ǫp ǫp ǫp = 1 + 1 − pr/2 pr/2 1 − pr/2 ! ! 2 − α ǫp ǫp ǫp ǫp + 1 + + 1 2 r r/2 α( r 1) r/2 p 1 − p ··· p 2 − − p 2 − ! ! α+1 1 ǫp ǫp ǫp − = 1 1 1 r/2 + r r/2 1 − p − (α 1)( 2 1) − p !  p −  − !   For r = 4, the thing becomes critical: Let us look at it more speciﬁcally. 335 r(p 1) − is even now and so ǫ = 1. Hence 4 p 1 1 1 pα+1 γp = 1 − − p2 1 1 ! − p We go to the full singular series.

S 4(n) = γp = γ2 γp p p 3 Y Y≥ 1 1 1 pα+1 = γ 1 − 2 − p2 1 p 3 p 3 1 p Y≥ ! Y≥ − 39. Lecture 245

The product is convergent since 1 < . So p2 ∞ 1 P 1 1 p2 S (n) γ 1 − | 4 |≥ 2 − p2 1 p p n 1 p Y ! Y| − 1 2 1 γ 1 ≤ 2 − p2 1 p p n 1 p Y ! Y| − 1 1 p diverges to zero in the infinite product senses. So S 4(n) is not bounded.− S (n) could become very small if we keep the odd factors fixed 4 andQ introduce more even factors. S 4(n) is unbounded in both senses; it can be as large as we please or as small as zero. For r 5 we are again on the safe side. In this case the first term comes ≥ from Vpλ . We have

Vp S (n) ∼ 1 5 ± p5/2 ! 1 1 or C 1 + < S (n) < C 1 2 p2 5 1 − p2 ! ! Y Y 336 For r = 7 the situation is similar. For r = 6 the series again converges. So for r 5. ≥ 0 < C1 < S r(n) < C2 This is of importance in the application to our problem. We had r/2 π r 1 r/4 A (n) = n 2 − S (n) + O(n ) r Γ(r/2) r r r If r 5, 2 1 > 4 , and since S r(n) being bounded does not raise the order in the term,≥ − r/2 π r 1 A (n) ∼ n 2 − S (n) r Γ(r/2) r If, however, if r = 4, the sharpness of the analysis is lost. Both the ﬁrst factor and the error term are O(r) and S r(n) may contribute to a decrease in the ﬁrst term. If there are many odd factors for n, the main term is still good. But if there are many powers of 2, it would be completely submerged. For r = 4 the exact formula was given by Jacobi. 2 + 2 + 2 + 2 We shall consider also representation of n in the form an1 bn2 cn3 dn4 in which connection the Kloosherman sums appear. We shall also cast a glance at the meaning of the singular series in the sense of Siegel’e p-edic density. Lecture 40

Let us look at S r(n) a little more explicitly. 337

S r(n) = γ2(n)γ3(n)

r 0 (mod 4). ≡ In this case we need not bother about the sign of the Gaussian sums; the fourth power of the coeﬃcient becomes 1.

γ (n) = 1 + V (n) + V 2 (n) + 2 2 2 ··· which is a ﬁnite sum. If 2 ∤ n, then γ (n) = 1. If2 n, n = 2αn ,2 ∤ n , then 2 | ′ ′

V2(n) = 0

( )r/4 − . if λ < α + 1; (λ 1) r 1 2 − ( 2 − )  r/4 λ =  ( ) V2 (n)  − , if λ = α + 1;  (λ 1)( r 1) − 2 − 2 −   + 0, if λ > α 1   So 

= + r/4 1 + 1 + + 1 1 γ2(n) 1 ( ) r r r 2 1 2( 1) (α 1)( 1) α r − 2 − 2 2 − ··· 2 − 2 − − 2 1   2 −  α n  r  ( ) 2µ  = 1 + ( ) 4  − ,  µ r − = 2 1 Xµ 1 2 − 246 40. Lecture 247

if 2α n (2α is the highest power of 2 dividing n). If2 ∤ n, γ (n) = 1. 338 || 2 1 1 1 γ (n) = 1 1 + , pα n p r/2 r/2 1 r − p − p ··· α( 2 1) || !  − p −   1 1 1 S (n) = γ (n) 1 1  + + r 2 r/2 r/2 1 r − p − p ··· α( 2 1) p 3 ! p n,p odd  − p −  Y≥ | Y   = γ (n)P P (n),   2 1 · 2   where P1 is a ﬁxed factor and

= 1 + + 1 P2(n) 1 r 1 α r 1 − p 2 ··· ( 2 ) p n,p odd  − p −  | Y   =  1  r . 2 1 d n,d odd d − | X 1 1 − 1 P = 1 1 1 − 2r/2 − pr/2 p 2 ! Y≥ ! 2r/2 1 = . 2r/2 1 × r − ζ 2 It is known (vide: Whittaker & Watson) that

2k k 1 (2π) B2k ζ(2k) = ( ) − , k 1, − 2(2k)! ≥ where B2k are the Bernoulli numbers. 339

1 k 1 B = , B = B = B = = 0; B , 0; sgnB = ( ) − 1 −2 3 5 7 ··· 2k 2k − ! r/2 2 r ! = 2 2 P1 r/2 r/2 2 1 × (2π) Br/2 − | | So for r > 4, the principal term

r/2 π r Ar(n) ∼ n 2 1 S n(n) Γ r − 2

= Cr(n), 40. Lecture 248

say, where

r/2 r/2 2 r ! π 2 2 r 1 1 C (n) = n 2 γ (n) r r/2 r/2 − 2 r 1 Γ(r/2) 2 1 (2π) B 2 r/2 d n,dodd d − − | | |X (a divisor sum! which is interesting, but not surprising, because the Jacobi 340 formula contains it).

r r 1 = 2 1 Cr(n) r n − r 2 1 2 1 2 − Br/2 d n,d odd d − | | | X r 0 (mod 8) ≡

r 1 1 n 2 − γ (n) 2 · dr/2 1 d n,d odd − | X r 1 1 1 1 1 = n 2 1 + + + . − r 1 (α 1)( r 1) r r 1 2 ··· 2 − α( 2 1) 2  2 − 2 − − 2 − d n,d odd d −   | X   n  r  ( ) δ  = 2 1   n −  −r , if n is even;  2 1 δ n δ − X| r 1 2 1 n − r , if n is odd. 2 1 δ n δ − X| So for any n,

n r 1 1 r 1 n ( ) δ n 2 γ (n) = n 2 ( ) − 2 r 1 − −r 1 2 − 2 d n,dodd d − δ n δ − |X X| r 1 n 2 = ( )n ( )n/δ − − − δ δ n X| n t r 1 = ( ) ( ) t 2 − − − t n X| So 341

n t r 1 C (n) = Q ( ) ( ) t 2 − , r r − − t n X| 40. Lecture 249

r here Q = r r 1 2 2 − B r | 2 | This is exactly what appears for r = 4 in the Jacobi formula. r 4 (mod 8) ≡

r 1 2 1 n − γ2(n) r 2 1 d n,dodd d − |X r 1 1 1 1 1 = n 2 1 + , d odd − r 1 (α 1)( r 1) r r 1 − 2 −···− 2 α( 2 1) 2 2 − 2 − − 2 − d n d − ! X| δ+ n +1 r ( ) δ = 2 1 n − − r 2 1 δ n δ − X| r 1 δ+ n +1 n 2 − = ( ) δ − δ δ n X| n +t+1 r 1 = ( ) t t 2 − , if n is even; − t n X| r 1 t 2 − , if n is odd; t n X| or in either case 342

n t+ n +1 r 1 ( ) ( ) t t 2 − − − t n X| n n +t 1 r 1 So C (n) = ( ) Q ( ) t − t 2 − r − r − t n X| n t+ r ( n +1) r 1 A (n) ∼ Q ( ) ( ) 4 t t 2 − ; r n − − t n X| r where A = r r 1 2 2 B / − | r 2| The Bernoulli numbers are all rational numbers and we can show that r/2 2k 2(2 1)Br/2 is an odd integral i.e., 2(2 1)B2k(k integral) is an all inte- 343 ger. Suppose− q is an odd prime; then, by Fermat’s− theorem,

q 1 2 − 1( mod q) ≡ Let (q 1) 2k. Then − | 22k 1 (mod q) ≡ 40. Lecture 250

22k 1 0 (mod q) − ≡ We now appeal to the non-Steadt-Clausen theorem, which is a beautiful theorem describing fully the denominators of the Bernoulli numbers: 1 B = G 2k 2k − p (p 1) 2k X− |

where G2k is an integer. 1 ∴ (22k 1)B = (22k 1)G (22k 1) − 2k − 2k − − p (p 1) 2k X− | 1 = integer + integer 2

2k So 2(2 1)B2k is an odd integer. Let us obtain− some specimens of

= 2r Qr r/2 (2(2 1) B / ) − | r 2| 32 A = 8, Q = 16, Q = 8, Q = , 4 8 12 16 17 8 16 8 64 Q = , Q = , Q = , q = 20 31 24 691 28 5461 32 929569

The conspicuous prime 691 appears in connection with the representation 344 as the sum 24 squares; it has to do with η24. Can Ar(n) be exactly equal to the asymptotic expression? (as for r = 4). A4(n) = C4(n), A8(n) = C8(n). From Q16 on wards, A16(n) , C16(n). This is because Q16 has an odd prime factor in the denominator. Suppose p divides the denominator. Then the fraction produced by Q16 cannot be destroyed by the other factor and Cr(n) is not always an integer. If p n. the numerator of C (pα) is congruent to 1 mod p. | r ± Lecture 41

It might be of interest to take Cr(n), the main term in the formula for Ar(n) and 345 make some remarks about it. r r n 1 n+d+ +1 d 2 − C (n) = Q ( ) 4 ( d ) r r − d n X| Let us form the generating function

∞ n Hr(x) = 1 + Cr(n)x ; = Xn 1 this will give a sort of partial fraction decomposition. In the case where r 0 (mod 8), it is simpler: ≡

∞ n n+d r 1 H (x) = 1 + Q x ( ) d 2 − r r − n=1 d n X X| ∞ n d r 1 = 1 + Q ( x) ( ) d 2 − r − − n=1 d n X X| ∞ d r 1 ∞ qd = 1 + Q ( ) d 2 − ( x) r − − = q=1 Xd 1 X d ∞ r ( x) = + d 2 1 1 Qn ( ) d − − d = − 1 ( x) Xd 1 − − d ∞ r 1 x = 1 + Q d 2 − r 1 ( x)d d=1 − − X 2 3 1.x r 1 x r 1 x = 1 + Q + 2 2 − + 3 2 − + r 1 + x 1 x2 1 + x3 ··· ( − ) 251 41. Lecture 252

This is a Lambert Series. Replacing x by eπiτ, it becomes

πiτ 2πiτ e r 1 e 1 + Q + 2 2 − + r 1 + eπiτ 1 e2πiτ ··· ( − ) The series above can be transformed into an Eisen stein series. If r is taken 346 to be 8, it is actually the 8th power of the V function Next, take r 4 (mod 8) − ≡

∞ n n+d+ n +1 r 1 G (x) = 1 + Q x ( ) d d 2 − r r − n=1 d n X X| ∞ d r 1 n n/d = 1 Q ( ) d 2 − ( x) ( ) − r − − − d=1 d n X X| ∞ d r 1 ∞ qd q = 1 Qr ( ) d 2 − ( x) ( ) − = − = − − Xd 1 Xq 1 d ∞ d r 1 ( x) = 1 + Q ( ) d 2 − − r + d = − 1 ( x) Xd 1 − d ∞ r 1 x = 1 + Q d 2 − r 1 ( x)d d=1 − − X 2 3 1 x r 1 x r 1 x = 1 + Q · + 2 2 − + 3 2 − + r 1 x 1 + x2 1 x3 ··· ( − − ) This is again a Lambert Series. This shows that a V power has to do with Lambert series which appears as an evaluation of certain− Eisenstein series not that they are identical. We now go to something quite diﬀerent. We had for r 5, ≥ r/2 π r 1 Ar(n) ∼ n 2 − S r(n)(*) Γ r 2

This comes out as a nice formula. Now could we not make some sense out 347 of this formula? What is its inner meaning? We shall show that the ﬁrst factor r/2 r 1 π /Γ(r/2) n 2 − gives the average value of the number of representations of n as the sum of r squares; the second factor also is an average, in the p-adic measurement. We shall show that

r/2 π r 1 A (n) ∼ n 2 − r Γ(r/2) n x n x X≤ X≤ 41. Lecture 253

So for each individual n, S r(n) gives the deviation of Ar(n) from r/2 r 1 π /Γ(r/2) n 2 − ; but on the average there is no deviation. Let us ﬁrst look at n x Ar(n). ≤

P ar(n) = 1 n x n x m2+ m2=n X≤ X≤ X1 ··· r = 1 , m2+ +m2 x X1 ··· r ≤ which is the number of lattice-points in the r-sphere with centre at the origin and radius √x, and so is proportional asymptotically to a certain volume (because the point lattice has cells or volume 1 and to each points belongs a cell). So this is roughly the volume of the sphere of radius √x which is

dx dx ··· 1 ··· r Zx2+ +x2Z x 1 ··· n≤ πr/2 = xr/2 Γ(r/2) 348 The diﬀerence will not be zero but of the order of magnitude of the surface of the sphere, i.e., O xr/2 1 Now consider the other− side. r/2 r/2 x π r 1 π r 1 n 2 − ∼ V 2 − dV Γ(r/2) Γ(r/2) n x 0 ≤ Z X r/2 r/2 = π x Γ r + 2 1

So the ﬁrst factor on the right of (*) gives the average. S r(n) has to be adjusted. S r(n) is also, surprisingly, an average. It was deﬁned as S (n) = γ (n)γ (n)γ (n) γ (n) , r 2 3 5 ··· p ··· and γp(n) in turn was given by 349

∞ γp(n) = 1 + Vpλ (n) = Xλ 1 ∞ h h 1 λ r 2πi λ n 1 λ r 2πi λ n = 1 + G(h, p ) e− p G(h, p ) e− p pλr pλr λ=1 h mod pλ h mod pλ X Xp∤h Xp∤h 41. Lecture 254

1 2πi h ℓ 2πi h ℓ = pλ 1 pλ 2 λr e e p λ  λ λ h mod p ℓ1 mod p ℓ2 mod p Xp∤h  X X   2πi h ℓr2 2πi h n e pλ e− pλ ℓ mod pλ  r X  h  1 2πi 2 2 = pλ + +  λr e (ℓ1 ℓr n) p λ λ ··· − ℓ1,...,ℓ4Xmod p h Xmod p = 1 λr p λ ℓ1,...,ℓrXmod p s 2+ + 2 t 2+ + 2 2πi λ (ℓ1 ℓr n) 2πi λ 1 (ℓ1 ℓr n) e p ··· − e p − ··· −  λ − λ 1  s mod p t mod p −   X X  1 2πi s (ℓ2+ +ℓ2 n) =  pλ 1 ··· r −  λr e  p λ λ ℓ1,...ℓnXmod p s Xmod p t 2+ + 2 2πi λ 1 (ℓ1 ℓ2 n) e p − ··· − − λ 1 λ 1 ℓ1,...,ℓr Xmod p − t modXp − = Wpλ (n) Wpλ 1(n), say, − − ∴ 1 + V (n) + V 2 (n) + + V λ (n) = W λ (n) γ (n) p p ··· p p → p So for λ large enough Wpλ (n) = Vpλ (n): the partial sums get identical. The 350 value of λ for which this occurs depends on the structure of n, on how many primes that speciﬁc n contains. Now

s 2+ + 2 2πi (ℓ ℓr n) λ e pλ 1 ··· − = 0 or p X pλ ∴ λ = Wp (n) λr p λ ℓ1,...,ℓr mod p ℓ2+ +ℓ2 Xn (mod pλ) 1 ··· r ≡ The sum on the right gives the number of times the congruence 2 2 λ ℓ + + ℓ n (mod p ) can be solved, N λ (n), say. Then 1 ··· r ≡ p 1 λ = λ Wp (n) λ(r 1) Np (n) p − We have therefore divided the number of solutions of the congruence by λ(r 1) λ λr p − . Now how many ℓ1,...,ℓr mod p do we have? There are p possibilities discarding n. n is one of the numbers modulo pλ. So dividing by pr, the 41. Lecture 255

λ λ(r 1) Np (n) average number of possibilities is p − . Hence λ(r 1) is the average density p − λ modulo p of the solution of the congruence. And since the Wpλ (n) eventually becomes γp(n), each factor γp(n) acquires a density interpretation, viz. the p-adic density of the lattice points modulo pλ. Lecture 42

The error term in the formula for the number of representations of n as the 351 sum of r squares, r 5, was O(nr/4). For r = 4 this did not suffice. We shall therefore study the≥ problem by Kloosterman’s method and find out what = 2 + 2 + 2 + 2 happens when we want to decompose n in the form n n1 n2 n3 n4. 1 We shall see that we can diminish the order in the error term by nearly 18 . When Kloosterman did this for the first time (Act a Mathematic as 1927) he took a slightly more general problem, that of representing n in the form n = 2 + 2 + 2 + 2 an1 b2 cn3 dn4, a, b, c, d integers. This works nicely; we get the singular series and an error term which is smaller than before. The difficult not will be about the arithmetical interpretation. The singular series willnowbea difficult phenomenon; we shall have multiplicativity, but the interpretation of the factors γp becomes complicated. We shall content ourselves with the analytical power of the discussion. The generating function which will have to be discussed is quite clear:

F(x) =Θ(xa) Θ(xb) Θ(xc) Θ(xd)

∞ 2 where Θ(x) = xn n= X−∞ And we will have 1 Θ(xa)Θ(xb)Θ(xc)Θ(xd) A (n) = dx 4 2πi xn+1 ZC and the analysis goes on as before with Farey series. We are here representing n by a positive definite quadratic form which is a diagonal form. Let us make the problem more general. Let us represent n by a positive definite form with integral coefficients. 352

256 42. Lecture 257

(We could very well unsedes also the ‘semi-integral’ case). Let S be a positive deﬁnite integral symmetric matrix and x a column vector with elements x1, x2,... xr in r-space. x′ is the transposed row-vector. x′Sx is a quadratic form in r variables. The question is how often can we express an integer n by integer vectors with respect to this quadratic form in r variables. The generating function to be studied this time is

n′S n Fr(x) = x , x < 1, n | | X the summation over all integral vectors n. Convergence is easily assured by positive definiteness. Indeed 2 2 x′Sx C(x + + x ), C > 0 ≥ 1 ··· r For x′Sx has a minimum C > 0 on x = 1 by positive definiteness; the in- | | c(n2+ +n2) equality follows from the homogeneityof the quadratic form. And x 1 ··· r is trivially a product of convergent series. In a later paper (Hamburger Abhandlungen, 1927) KloostermaP n, on the advice of Hecke, took up a more general problem. This would require a little more preparation on modular forms. The generating function will now be r a modular form of dimension 2 of a certain ‘stafe’; so we have to discuss modular forms not only with respect− to the full modular group, but also the substitutions a b 1 0 (mod N), c d ≡ 0 1     (N will the ‘stafe’) which from a subgroup  finite index in the modular group. 353     Kloosterman’s work goes through for all modular forms of this sort, but we should want generalisations of η(τ) and V (τ). To do this we need a good deal of Heeke’s theory about Eisenstein series of higher stafe of the type: 1 r (m1 + m2τ) m1 a (mod N) m ≡b X(mod N) 2≡ r which is a modular form of dimension 2 and stafe N. These were investigated by Hecke in a famous paper (Hamburger− Abhandlungen 1927). Kloosterman could carry out his theory for these also. We shall, however, compromise on the quadratic form. We had the generating function

n′S n Fr(x) = x , x < 1, n | | X 42. Lecture 258

∞ n = 1 + Ar(n)x . = Xn 1 Fr(x) is a modular form. This can be seen directly by the transformation formulae. Let us start with Kloosterman’s method and see what happens. The problem is to get 1 F (x) A (n) = n dx r 2πi xn+1 ZC At a certain moment later on we shall need a greater knowledge of Fr(x) Let us carry out the Farey dissection:

2πi h 2πz 2πi h 2π(δ iϕ) x = e k − = e k − N − V hk′′ 2πi h n 2πi h 2πz 2πnzdϕ Ar(n) = e− k Fr(e k − )e 0 h

1 1 with (h, k) = 1, V ′ = , V ′′ = where in the Farey situation, 354 hk k(k1+k) hk k(k+k2) h1 < h < h2 . The reﬁnement of Kloosterman consists in not merely making k1 k k2 the rough remark that 1 1 V ′ , V ′′ , 2kN ≤ hk hk ≤ k(N + 1) but in a ﬁner following up of the number theoretical determination of the adjacent fractions. We have

h k hk = 1, hk h k = 1; 1 − 1 − 2 − 2 − i.e., hk 1 mod k, hk 1 mod k 1 ≡ 2 ≡− h k is given. What we are worried about is, how long is its environment. k1 and k2 are given as solutions of certain congruences. We have the habit of calling h′ a number such that

hh′ 1 (mod k); soletuswrite ≡− k h′ (mod k), k h′ (mod k) 1 ≡− 2 ≡

So we know in which residue class modulo kk1 and k2 have to lie. k1 + k, 355 being the denominator of a mediant, had to exced N. N < k + k N + k, or 1 ≤ N k < k1 N. So k1 has a span of size k. This along with k1 h mod k determines− k≤ completely. Similarly, for k , N k < k N So≡ −there is no 1 2 − 2 ≤ 42. Lecture 259

V V uncertainty at all about hk′ , hk′′; and we could single them out if we insisted on that. For example, let h = 5 , N = 12; what are the neighbours? h1 < 5/9 < h2 . k 9 k1 k2 First determine h′. 5h′ 1 (mod 9) or h′ = 7. Then 12 9 < k1 12 and k 7 (mod 9), so k≡= − 11. Similarly 3 < k 12, k− 7 (mod≤ 9) so 1 ≡ − 1 2 ≤ 2 ≡ k2 = 7. We need only k1 and k2; but for our own enjoyment let us calculate h1 and h2. h, 5 5 h = 1, 2 = 1, 11 9 − 9 7 −

= = 6 5 4 or h1 6, h2 4, so that we have 11 < 9 < 7 as adjacent fractions in the Farey series of order 12. We do not need to display the whole Farey series. Now utilise this in the following way.

1 + k(k k2) h h ′ 2πi n 2πi 2πz 2πnz Ar(n) = e k Fr e k − e dϕ

o h

Kloosterman does the following investigation. In any case we are sure that k1, k2 can at most become N. If we take k1 and k2 big we have a small interval of integration. Since k + k < k + 1 + k < < N + k, 1 1 ··· k + k < k + 1 + k < < N + k, 2 2 ··· 1 1 1 1 > , > , k1 + k N + k k2 + k N + k so that the interval of integration should be at least as big as the interval 356 1/k(k + N) to 1/k(k + N). This interval is always present whatever be k − 1 and k2. So Ar(n) is equal to the always present kernel

1 k(k+N) 2πi h n ′ e k ( )dϕ, ··· 0 h

1 1 + + + N 1 k(k ℓ) N 1 k(k ℓ 1) 2πi h n − 2πi h n − ′ e− k ( )dϕ e− k ( )dϕ = ··· = ··· 0 h

There is no doubt about the integrals. The limits are all well-deﬁned. This will help us to appraise certain roots of unity closely-by the Kloosterman sums. 357 We shall now return to the integrand; that is a V -function and requires the usual V treatment. Consider the r-fold V -series:

πtn′S n Θ(t) = e− , Ret > 0. n X Modify this slightly by introducing a vector α of real numbers; α′ = (α1,...,αr). Let

πt(n′+α′)S (n+α) Θ(t; α1,...,αr) = e− n X

This is periodic in α j, of period 1, and so permits a Fourier expansion. The convergence is so good that the function is analytic in each α j and so we are sure that it is equal to the sum

C(m)e2πim′α m X where C(m) is the Fourier coeﬃcient:

1 1 2πim′β C(m) = Θ(t; β1,...,βr)e− dβ1,... dβr 0 ··· 0 Z 1 Z 1 πt(n′+β′)S (n+β) 2πim′β = e− e− dβ1,..., dβr 0 ··· 0 n Z Z X 1 1 πt(n′+β′)S (n+β) 2πim′(n+β) = e− e− dβ1,..., dβr 0 ··· 0 n Z Z X which is an integral over the unit cube W, and so on translation with respect to 358 the vector n, becomes

V V V πt( S ) 2πim′ V V e− e− d 1 d r n ··· ··· X Z W+nZ (the exchange of integration and summation orders being trivial)

∞ ∞ πtV ′DV 2πim′V = e− e− dV ... dV . ··· 1 r Z−∞ Z−∞ Lecture 43

Let us return to the generalised theta-formula: 359

πt(n′+α)S (n+α) Θ(t; α1,...αr) = e− n X = c(m)e2πim′α m X where ∞ πtV S V 2πim V c(m) = e− ′ e− ′ dV ... dV ··· 1 r Z Z −∞ To get this into shape, consider the quadratic complement

π 1 1 π 1 (tV ′ + im′S − )S (tV + iS − m) = πtV ′S V πiV ′m πim′V + m′S − m − t − − − t

Since m′V = V ′m,

∞ π m S 1m π (tV +im S 1)S (tV +iS 1m)dV ...dV c(m) = e− t ′ − e− t ′ ′ − − 1 r ··· Z Z −∞ ∞ π 1 i 1 i 1 m S m π( √tV ′+ m′S − )S ( √tV + S − m) = e− t ′ − e− √t √t dV dV ··· 1 ··· r Z Z −∞ Put √tV = w and µ = 1 m S 1. Then 360 √t ′ −

π 1 t m′S − m ∞ e− π(w′+iµ′)S (w+iµ) c(m) = e− dw1 dwr ( √t)r ··· ··· Z Z −∞ 261 43. Lecture 262

Since every positive deﬁnite quadratic form may be turned into a sum of squares, we can put S = A′A, so that the exponent in the integrand become π(w A + iµA )(Aw + iAµ); and writing Aw = z, we have − ′ ′ ′

π 1 t m′S − m ∞ e− π(z +iV )(z+iV ) dz1 ... dzr c(m) = e− ′ ′ ( √t)r ··· A Z Z | | −∞ 2 where V = µA, and A = determinant of A. Let D = A = S , z′ = (z1,..., zr). Then | | | | | |

π 1 m′S − m r e− t ∞ π(z +iV )2 c(m) = e− j j dz D1/2tr/2 j j=1 Y Z−∞ π 1 m′S − m r e− t ∞ πz2 = e− dz D1/2tr/2 Z−∞ ! π 1 e t m S m = − ′ − , D1/2tr/2 the last factor being unity. So we have ultimately

1 π m S 1m 2πim α Θ α ,...,α = t ′ − ′ (t; 1 r) 1/2 r/2 e− e D t m X

Let us nowwe back to our study of Ar(n). We had integrals with now limits 361 which were the special feature of the Kloosterman method.

1 + k(k N) N 1 N 1 2πi h n 2πi h 2πz 2πnz − − Ar(n) = ′ e− k Fr e k − e dz + + 0 h

∞ n′S n n Fr(x) = x = 1 + Ar(n)x n = X Xn 1 2πi h 2πz (2πi h 2πz) Fr e k − = e k − n′Sn n X 2πi h n S n 2πzn S n = e k ′ e− ′ n X 43. Lecture 263

n is of interest only modulo k, so put

n = kq + ℓ,ℓ = (ℓ ,...,ℓ ), 0 ℓ < k. 1 n ≤ j So dismissing multiples of k,

h h 2 ℓ′ ℓ 2πi 2πz 2πi ℓ′S ℓ′ 2πzk (q′+ )S q+ Fr(e k − ) = e k e− k k , q ℓ Xmod k X and applying the transformation formula we derived earlier, with t = 2zk2 and = 1 α k ℓ, this becomes

h π 1 ℓ 1 2πi ℓ S ℓ m′S − m 2πim ′ e k ′ e− 2zk2 e ′ k r r/2 r/2 √Dk e z m Xℓ X π 1 1 2 m′S − m = e− 2zk Tk(h, m), r r/2 √Dk (2z) m X on exchanging summations, where 362

πi 2 (hℓ′S ℓ+m′ℓ) Tk(h, m) = e k Xℓ

Tk(h, 0) will be the most important; the others we only estimate. We require a little more number theory for this. We cannot tolerate the presence of a both a quadratic form and a linear form in the exponent. There will be a common 1 denominator in m′S − m and that will have to be discussed. Lecture 44

We had 363

h π 1 2πi 2πz 1 m′S − m F (e k ) = e− 2zk2 T (h, m), r − r z r/2 1/2 k k (2 ) D m X 2πi (hℓ′S ℓ+m′ℓ) and Tk(h, m) = e k ℓ Xmod k 1 The common denominator in m′S − m will be at most D, the determinant; deﬁne k∗ and Dk by

kD = k (k, D) D = k∗D , (D , k) = 1, · · k k k so that Dk is D stripped of all its common divisors with k. Suppose ﬁrst that k is odd. Let ρ be a solution of the congruence

4hD ρ 1 (mod k∗) k ≡ h + 2πi k (ℓ′S ℓ 4Dk ρm′ℓ) Tk(h, m) = e ℓ Xmod k h 1 1 2 2 1 h 2πi (ℓ′+2D ρm S )S (ℓ+2D ρS m) (4D ρ m S m)2πi = e k k ′ − k − e− k ′ − k . ℓ Xmod k h 2 2 1 h 1 1 2πi 4D ρ m S m 2πi (ℓ′+2D ρm S )S (ℓ+2D ρS m) = e− k · k ′ − e k k ′ − k − ℓ Xmod k Dkρ 1 2π m′S − m = e− k Uk, say, (using the deﬁnition of ρ), where Uk = Uk(h, m) is periodic in m with 364 period (k, D); it is enough if we take this period to be D itself. So

D ρ h 1 π +2πi k m S 1m 2πi 2πz z 2 k ′ − Fr(e k − ) = Uk(h, s) e− 2 k kr(2z)r/2D1/2 s mod D m s (mod D) X ≡ X 264 44. Lecture 265

This is a linear combination of ﬁnitely many V -series of the form

1 xm′S − m m s (mod D) ≡ X x 1 The power series goes in powersof D because D remains silent inside. This is for k odd. For k even, deﬁne. σ by

hD σ 1 (mod4k∗) k ≡

h 2πi (4ℓ′S ℓ+4Dk σm′ℓ) Tk(h, m) = e 4k ℓ Xmod k 2πi h D2σ2m S 1m 2πi h (2ℓ +D σm S 1)S (2ℓ+D σS 1m) = e− 4k k ′ − e 4k ′ k ′ − k − ℓ Xmod k Dk 1 2πi σm′S − m = e− 4k Uk(h, m),

where Uk again has a certain periodicity; we can take the period to be 2D and 365 forget about the reﬁnement. So

h π + 2πi 1 2πi 2πz 1 2 4k Dk σ m′S − m F e k − = U (h, s) e− 2zk r kr(2z)r/2D1/2 k s mod 2D m s (mod 2D) X ≡ X

which is again a linear combination of theta-series with coeﬃcients Uk. Ob- serve that Tk and Uk diﬀer only by a purely imaginary quantity:

T (h, m) = U (h, m) , | k | | k |

and for m = 0, Tk(h, 0) = Uk(h, 0). We shall use as essential only those theta-series which are congruent to zero modulo D or 2D; and the rest will be thrown into the error term. Only these corresponding to o have a constant term. The general shape in both cases is 1 m′S − m Uk(h, s) = x s mod 2D m s (mod 2D) X ≡ X Lecture 45

We have to get a clear picture of that we are aiming at. We are discussing the 366 function under the integral sign. We get it as

2πi h 2πz 1 F e k − = U (h, s) r kr(2z)r/2D1/2 k s mod 2D X π + Dk σ 2 2πi 4k 1 e− 2k z m′S − m m s (mod 2D) ≡ X where k D = k∗Dk, (k, Dk) = 1. k is even; if k is odd the formula looks ﬁnitely many diﬀ· erent values. This most important fact we formulate as a lemma.

Lemma 1. For k even, Uk(h, s) depends only on h modulo 2D. This depends on a theorem on the behaviour of quadratic forms the equiv- alence of quadratic forms modulo a given number. This is a lemma of Siegel’s (Annals of Mathematics, 1935, 527-606). Let us recall that for k even

i 2π (h ℓ′ S ℓ +m′ ℓ) Tk(h, m) = e k ℓ Xmod k h 1 2πi Dkσm′S − m = e− 4k Uk(h, m)

Lemma 2. T (h, m) Ckr/2 | k |≤ We have

h h 2 2πi (ℓS ℓ+σm ℓ) 2πi (λ′ S λ +σm ℓ) T (h, m) = e k ′ e− k ′ | k | ℓ Xmod k λ Xmod k 266 45. Lecture 267

h 2πi (ℓ′ S ℓ λ S λ +σm (ℓ λ)) = e k − ′ ′ − , Xℓ,λ and since 367

ℓ′ S ℓ λ′ S λ = (ℓ′ λ′)S (ℓ + λ) + λ′ S ℓ ℓ′ S λ − − − = (ℓ′ λ′)S (ℓ + λ) + ℓ′ S ℓ ℓ′ S λ − − = (ℓ′ λ′)S (ℓ + λ), − this is equal to

h 2πi (ℓ′ λ′)(S (ℓ + λ)+σm) e k − Xℓ,λ 2πi h α (S (ℓ + λ)+σm) = e k ′ α mod k ℓ λ α mod k X − ≡X 2πi h α (S (2 λ +α)+σm) = e k ′ α mod k ℓ λ mod k X − ≡X 2πi h α (S α+σm) 2π h 2α S λ = e k ′ e k ′ α Xmod k λ Xmod k

If we write 2α′S = β′, the inner sum is 368

2πi h (β λ + +β λ ) 2 e k 1 1 ··· r r = k , if k β ,..., k β ; | 1 | r λ1,...λXr mod k 0 otherwise

2 So Tk(h, m) = 0 if at least one β is not divisible by k; otherwise it is equal to | | r 2πi h α (S α+σm) k e k ′ α Xmod k Writing S = (s jk), the system of congruences

i.e., T (h, m) 2r/2 S r/2kr/2. | k |≤ | | We shall now outline the main argument a little more skillfully, putting the 369 thing back on its track. Ar(n) is the sum of integrals over the ﬁner-prepared Farey arcs of Kloosterman:

1 k(k+N) 2πnz 1 2πi h n e A (n) = e k r r/2 1/2 ′ − r/2 2 D 0 h

1 m′S − m Θs(x) = x ; m s (mod 2D) ≡ X = S 0 + S 2 + S 1, say,

= S 00 + S 0m + S 20 + S 2m + S 10 + S 1m  m,0   m,0   m,0   X   X   X        in an obvious notation. Now treat  the things separately.  By inspection of Uk(h, m) we ﬁnd how it depends on h, it is only modulo 4 k∗. We have to reconcile Lemma 1 with this. This actual period therefore is neither 2D nor 4k∗ but the greatest common divisor

kD (2D, 4k∗) = 2(D, 2k∗) = 2 d, D k ! 2 2D = (DDk, 2kD) = (Dk, 2k) Dk Dk 2D 4D = or Dk Dk

So we have 370 45. Lecture 269

2D Corollary of Lemma 1. Uk(h, m) for k even depends on h only modulo = Dk , say. ∧ 1 l(k+N) 2πnz 1 2πi h n e S = e− k T (h, o) dϕ 00 2r/2D1/2 k zr/2 0 h

1 k(k+N) π 2 1 D − 2k z 1 2πi h n 2πi k σm S 1m e m′S − m S = e k U (h, m)e 4k ′ − dϕ om r/2 1/2 ′ − k r/2 2 D 0 h

h i 1 2πi 2π Dk σm′S − m Kk(n, m) = ′ e− k Uk(h, m)e 4k h mod k 1 P 2πiahn.4+2πiV σ = U (λ, m) e− 4k∗ , a k λ mod h λ (mod ) ∧ ≡ ∧ X h modX 4k∗

= V = k∗ 1 1 where 4k∗ ak, a 4D, and k Dkm− S − m We deﬁned σ by≤ hD σ 1 (mod4k∗) k ≡ Let DD¯ 1 (mod4k∗), hH¯ 1 (mod4k∗) k ≡ ≡ Then

2πiahn+2πiV σ 1 − K (n, m) = U (λ, m) e 4k∗ D Dk¯ k a k k λ mod h λ (mod ) ∧ ≡ ∧ X h modXk∗,(h,k∗)=1

1 2πi ¯ ( 4anh+V D¯ kh) = Uk(λ, m) ′ e 4k − a h λ (mod ) λ mod ≡ ∧ X ∧ h Pmod k∗ The inner sum here is a Kloosterman sum. It has essentially 4k∗ terms. A 372 trivial estimate of this would be O(k), and this is what we had tacitly assumed in the older method. The advantage here is, however, that they can be appraised letter. We shall not estimate them here but only quote the result as 45. Lecture 270

Lemma 3.

2π i (uh+νh¯) K (u, ν) = e k , k, hh¯ 1 (mod k) k ∧| ≡ h λ (mod ) ≡h Xmod k∧ 1 α+ǫ α = O k − (u, k)

There has been a lot of discussion about the size of the α in this formula. = 1 Kloosterman and Esternann proved that α 4 (Hamb, Ab. 1930), Salie’ that = = 1 α 1/3 and A.Weil that α 2 (P.N.A.S’, 48) Weil’s was a very complicated and deep method going into the zeta-functions of Artin type and the Riemann hypothesis for these functions. We thus save a good deal in the order of magnitude. The further S’s will be nearly similar; the complete Kloosterman sums will be replaced by sums with certain conditions.

1 + N k(k N) R π 1 α − 2k2z · D 1 α+ǫ (n, k) π m S 1m 1 e 4 ( ′ − D ) ϕ S om C k − r/2 e− − r/2 d | |≤ = k z k 1 Z1 | | X + − k(k N) Since R 1 z 1 on the Farey arc, the integrand is majorised by k2 ≥ k

π δN r/4 π δN 1 δ 2D 2 − 2D k2 δ2 +ϕ2 2 2 2 r/4 r/4 N − k2 δ +ϕ2 e | N k (δ + ϕ ) − = δ− e ( N ) | N | N k2(δ2 + ϕ2)  N  = O(nr/4)    √n π 1 1 r/4 1 α+ǫ α m′S − m S om Cn k − (n, k) e− 4 , | |≤ = k √n Xk 1 since the path of integration has a length of the order 1/k √n. Now summing 373 over all m , 0,

√n r 1 α+ǫ α S Cn 4 − 2 k− (n, k) om ≤ m,0 k=1 X X r 1 α α+ǫ < Cn 4 − 2 d (dt)−

d n dt √n X| X≤ r 1 ǫ α+ǫ = Cn 4 − 2 d t− d n t √n X| X≤ d 45. Lecture 271

1 α+ǫ r 1 ǫ √n − < Cn 4 − 2 d d d n X| ! r α + ǫ α 1 = Cn 4 − 2 2 d − , d n X| and since the number of divisors of n is O(nǫ/2). This is 374

r α + ǫ + ǫ = Cn 4 − 2 2 2 r α +ǫ = Cn 4 − 2

Improving α has been the feature of many investigations. Lecture 46

All the other sums that we have to estimate behave some what similarly. We 375 take as specimen S 20.

1 k(k+ℓ) N 1 2πnz 1 2πi h n 1 − e = k , ϕ S 20 1/2 r/2 ′ e− Tk(h o) r r/2 d D 2 o h

The original interval for k2 was bigger: N k < k2 N. Now the full interval is not permissible, i.e., we have admitted− not all residues≤ modulo k, but only a part of these, and the N may lie in two adjacent classes of residues. Here we have a new type of sum of interest. We know how to discuss Tk; h plays a role there. The sums we have now get are

2πi h n Tk(λ, o) ′ e− k h λ (mod ) λ mod ≡ ∧ ∧ N k

The inner sum is an incomplete Ramanujan sum, with restriction on k2 376 implying (see lecture 45) actually a restriction on h! The Kloosterman sums are a little more general:

2π i (uh+V h¯) e k hh¯ 1 (mod k) ≡ X Our present sums are incomplete Kloosterman sums (with V = o and u =

272 46. Lecture 273

1), and the interesting fact is that they also permit the same appraisal, viz.

r/2 1 α+ǫ α O k k − (k, n) From there on things go just as smoothly as before.

1 √n k(N+1) r/2 1 α+ǫ α dϕ S = O k− k − (k, n) 2o  2 + 2 r/4   = (δN ϕ )   k 1 Z1  X +   k(k N)    and here for convergence of the integral we want r 3. This would give again the old order. Similar estimates hold for the other pieces:≥ α S = O nr/4 + ǫ 2m 2 m,o − X (The incomplete Kloosterman sums here are actually incomplete Ramanu- jan sums and so we may got a slightly better estimate; but this is of no consequence as the other terms have a higher order). 377 We then have

r/4 α/2 ǫ 1 A (n) = S + O n − − , α = . r oo 2 Let us look at S oo. It is classical, but not quite what we like it to be.

1 k(k+N) 2πnz 1 2πi h n Tk(h, o) e r/4 α/2+ǫ S = ′ e− k dϕ + O n − oo D1/22r/2 kr zr/2 o h

√n ∞ 1 Hk(n) 2πnz r/4 α/2+ǫ dϕ + O n − , D1/22r/2 kr zr/2 k=1 Z X −∞ h 2πi nTk(h,o) with Hk(n) = e− k Xh r/2 1 α+ǫ α = O k k − (k, n) , thereby adding an error term of order

√n ∞ r +1 α+ǫ α dϕ O k− 2 − (k, n)  (δ2 + ϕ2)r/4   k=1 N  X Z1   kN      46. Lecture 274

Now

∞ ∞ dϕ dϕ 1 r/2 = δ − 2 + 2 r/4 2 r/4 N (δN ϕ ) ϕ Z1 Z1 1 + kN kN δN ∞ r 1 dψ = O n 2 −  (1 + ϕ2)r/4   ZN   k      = 2  N  with ψ N ϕ. ψ is never smaller than 1 as k > 1. So we can drop 1 in the 378 denominator without committing any error in the order of magnitude. So this gives r ∞ dψ 2 1 O n − / N ψr 2  Z k    and the integral converging for r 3, it is equal to ≥  r + 2 1 r 1 √n − r 1 r 1 O n 2 − = O n 4 − 2 k 2 −  k  !   Hence our new error term is 

√n α+ǫ α r 1 r/4 α/2+ǫ O k− (n, k) n 4 − 2 = O n −  k=1  X    which is what has already appeared.  We than have on writing 2πnz = ω,

c+i √n ∞ ω 1 Hk(n) 1 r 1 e r/4 α/2+ǫ = π 2 ω + , Ar(n) 1/2 r/2 r (2 n) − r/2 d O n − D 2 = k i ω k 1 c Zi X − ∞ and the integral being the Hankel integral for the gamma-function, 379

r/2 r/2 √n (2π) n 1 Hk(n) 1 r/4 α/2+ǫ A (n) = − + O n − r D1/22r/2 kr Γ(r/2) k=1 X r/2 π r/2 1 ∞ Hk(n) r/4 α/2+ǫ = n − + O n − Γ r 1/2 kr D k=1 2 X 46. Lecture 275

r +1 α+ǫ α r 1 ∞ k 2 − (k, n) + O n 2 − kr  k= √n+1   X    This new error term is  

r 1 α 1 r α+ǫ r 1 1+ǫ r/2 r/2 O n 2 d (qd) 2 = O n 2 d q  − − −   − − −   d n q> √n   d n q> √n   X| Xd   X| Xd      (This is because for the Ramanujan sum we have) 

2πi h n k ′ e− k = dµ d h mod k d (k,n) X X| ! = O (k, m) 1 = O (k, n)1+ǫ ;  d (k,n)   X|     1 α+ǫ α and then we use the old appraisal (k − (k, n) with α = 1 + ǫ). So we have 380

r/2+1 r 1 1+ǫ r/2 √n − r 1 ǫ r/4 1/2+2ǫ O n 2 − d − = O n 4 − 2 d = O n − d  d n !   d n   X|   X|      This is of smaller order than the old error term. So we have our ﬁnal result:

r/2 π r/2 1 ∞ Hk(n) r/4 α/2+ǫ A (n) = n − + O n − ; r Γ(r/2)D1/2 kr k=1 X the singular series plus the error term. What remains to be shown is that the singular series again enjoys the multiplicative property: = Hk1k2 (n) Hk1 (n)Hk2 (n) We shall then have it as the product

γp p Y H (n) Hp2 (n) where γ = 1 + k + + p pr p2r ···

The arithmetical interpretation now becomes diﬃcult, because all the prop- 381 erties that the quadratic form may have will have to show up. One or other of the factors γp may be zero in which case we have no representation. 46. Lecture 276

We should like to throw some light on the Kloosterman sums. We take for granted the estimate

2π i (uh+V h¯) 1 α+ǫ α ′ e k = O k − (k, u) h mod k · hh¯ 1 (mod k) ≡ P = 1 Kloosterman and Esterman (Hamburger Abhandlungen Vol.7) proved α 4 ; = 1 Salie’ (Math. Zeit., vol. 36) proved α 3 . Using the multiplicativity, in a certain sense, of the sums, Salie’ could prove that if k = pβ, p prime and β 2, = 1 ffi ≥ then α 2 but he could not provethis in the other cases. The di cult case was that of 2πi/p(uh+V h¯) ′ e . h mod p P + For this nothing better than O p2/3 ǫ(p, u)1/3 could be obtained; and it de- fied all efforts until A.Weil proved α = 1/2 in all cases by using deep methods (Proc. Nat. Acad. Sc.1948). Further application of the Kloosterman sums offer no difficulty. The (generalised) Kloosterman sums are symmetrical in u and V , for

2πi (uh+V h¯) 2πi (uh¯+V h) ′ e k = ′ e k h λ( ) h¯ λ¯( ) h ≡Xmod∧ k h ≡Xmod∧ k

since (λ, ) = 1, h λ (mod ) and hh¯ 1 (mod ) imply h¯ λ¯ (mod ) 382 and λλ¯ ∧1 (mod ≡). The last we∧ can write≡ as ∧ ≡ ∧ ≡ ∧ 2πi (uh+V h¯) ′ g(h¯)e k , h mod k where g(m) is the periodic functionP deﬁned as

1 if m λ¯ (mod ) g(m) = ≡ ∧  0 otherwise.   g(m) has therefore the ﬁnite Fourier expansion

= ∧ 2πi j m g(m) C je ∧ = Xj 1 The coeﬃcients c j can be calculated in the usual way:

1 2πiqλ¯ c = e − mq = 1, 2,..., q ∧ ∧ ∧ 46. Lecture 277

Substituting for Cq, the sum becomes

h¯ i λ¯ i + V + jk ¯ 2πij 2π (uh+V h¯) 1 2πij 2π k uh ( )h C ′ e e k = e− e ∧ j ∧ ∧ h mod k j mod ∧ j mod h mod k X ∧ P X ∧ X so that the generalised sum becomes a ﬁnite combination of undisturbed 1 α+ǫ α Kloosterman sums and so has the estimate O k − (k, u) ¯ This works just as well in the other case when there is an inequality on h. 383 2πi (uh+V h¯) 2πi (uh+V h¯) e k = f (h¯)e k h λ (mod ),h mod k h λ (mod ) ≡ aXh¯∧ b ≡h Xmod k∧ ≤ ≤ 1, 0 < m a, where f (m) = ≤  0, a < m k,  ≤ and f (m) is periodic modulo k.   Then k 2πi j m f (m) = c je k = Xj 1 where

2πij 2πi j(a+1)/k 1 e k e c = − − , j , k, j − 2πi j/k k 1 e− a − c = k k 2 c | j|≤ k sin π j/k The sum becomes 384

k 1 − 2πi (uh+(V + j)h¯) 2πi (uh+V h¯) c j ′ e k + ck ′ e k = h mod k h mod k j 1 h λ (mod ) h λ (mod ) X ≡ P ∧ ≡ P ∧ k 1 = 1 α+ǫ α + 1 − 1 O k − (h, k) 1 π j  k = sin   j 1 | k |  X  2 Since sin α ,   ≥ π   k 1 − j 1 π 1 2 π j 2 π j = sin ≤ 2 Xj 1 k X k 46. Lecture 278

1 = k = O(k log k) k 1 j j − X≤ 2 so that again the sum becomes

1 α+ǫ α O k − (k, u)

Kloostermanﬁrst discussed his method for a diagonal quadraticform. Later on he applied it to modular forms and for this he could derive on the investigations by Hecke comparing modular forms with Eisenstein series. In this case the theory becomes simpler: we can subtract suitable Eisenstein series and the principle term then becomes zero. The r-fold theta-series that we had are in fact modular forms.