RAMANUJAN S CLASS INVARIANTS, KRONECKER S LIMIT FORMULA, AND MODULAR EQUATIONS

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 39, Number 6, June 1997, Pages S )0173- RAMANUJAN S CLASS INVARIANTS, KRONECKER S LIMIT FORMULA, AND MODULAR EQUATIONS BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG In Memory of Jerry Keiper Abstract In his notebooks, Ramanujan gave the values of over 100 class invariants which he had calculated Many had been previously calculated by Heinrich Weber, but approximately half of them had not been heretofore determined G N Watson wrote several papers devoted to the calculation of class invariants, but his methods were not entirely rigorous Up until the past few years, eighteen of Ramanujan s class invariants remained to be verified Five were verified by the authors in a recent paper For the remaining class invariants, in each case, the associated imaginary quadratic field has class number, moreover there are two classes per genus The authors devised three methods to calculate these thirteen class invariants The first depends upon Kronecker s limit formula, the second employs modular equations, the third uses class field theory to make Watson s empirical method rigorous 1 Introduction So that we may define Ramanujan s class invariants, set a; q) 1 aq n ), q < 1, n0 11) χq) q;q ) If q exp π n),the two class invariants G n g n are defined by 1) G n : 1/ q 1/ χq) g n : 1/ q 1/ χ q) In the notation of Weber [36], G n : 1/ f n) g n : 1/ f 1 n) It is well known that G n g n are algebraic; for example, see Cox s book [11, p 1, Theorem 103; p 57, Theorem 117] At scattered places in his first notebook [3], Ramanujan recorded the values for 107 class invariants On pages 9 99 in his second notebook [3], Ramanujan gave a table of values for 77 class invariants, three of which are not found in the first notebook Since the second notebook is an enlarged revision of the first, it is unclear why Ramanujan failed to record 33 class invariants that he offered in the Received by the editors March 16, 1995, in revised form, November 1, Mathematics Subject Classification Primary 11R9; Secondary 11R0, 11R37, 11R, 11F7, 33D10 15 c 1997 American Mathematical Society License or copyright restrictions may apply to redistribution; see

2 16 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG first notebook By the time Ramanujan wrote his paper [], [, pp 3 39], he was aware of Weber s work [36], so his table of 6 class invariants in [] does not contain any that are found in Weber s book [36] Except for G 35 G 363, all of the remaining values are found in Ramanujan s notebooks In two papers [30], [31], G N Watson proved of Ramanujan s class invariants from [] In the first [30], Watson devised an empirical process to calculate 1 of the invariants, while in the second [31], he employed modular equations to prove 10 invariants In another paper [9], Watson established Ramanujan s value for G 1353, communicated by him in his first letter to Hardy [, p xxix], also stated in [] Watson wrote four further papers [3] [35] on the calculation of class invariants Among the dozens of invariants calculated by Watson in these papers were three previously unproved invariants found in Ramanujan s paper [], as well as 11 invariants of Ramanujan that had been previously verified Thus, after Watson s work, 1 invariants of Ramanujan from his paper notebooks [3] remained to be verified The authors established five of these invariants in [] For each of these five values, n is a multiple of 9, proofs were effected by formulas relating G 9n with G n g 9n with g n As a bonus, the latter two formulas led to closed form evaluations of Ramanujan s cubic continued fraction at the arguments ± exp π n) The purpose of this paper is to establish the remaining 13 values, each for G n,n 65, 69, 77, 11, 15, 05, 13, 17, 65, 301, 5, 505, 553, claimed by Ramanujan Quite remarkably, the class number for each of the 13 imaginary quadratic fields Q n) equals Moreover, there are precisely two classes per genus in each case Our first proofs employ the Kronecker limit formula, which is used to find representations for certain products of Dedekind eta functions in terms of fundamental units; see Theorems 31, 3, 5 Each of the 13 values of n is a product of a small prime 3, 5, or 7) a larger prime Thus, our proofs, given in Sections, also crucially employ certain modular equations of Ramanujan of degrees 3, 5, 7 It is highly unlikely that Ramanujan was familiar with the Kronecker limit formula the arithmetic of quadratic fields, so our proofs certainly are not those found by Ramanujan However, Ramanujan obviously discerned some unique arithmetical properties in these instances, it would be fascinating to discover Ramanujan s approach To make clearer the connection between modular equations class invariants, we first give a precise definition of a modular equation Let K, K,L, L denote complete elliptic integrals of the first kind associated with the moduli k, k : 1 k,l, l : 1 l, respectively, where 0 <k,l<1suppose that 13) n K K L L for some positive rational number n If n is a positive integer, a relation between k l induced by 13) is called a modular equation of degree n Following Ramanujan, set α k β l We often say that β has degree n over α As usual, in the theory of elliptic functions, set 1) q : exp πk /K) License or copyright restrictions may apply to redistribution; see

3 RAMANUJAN S CLASS INVARIANTS 17 Note that if K /K n where n is not necessarily the degree of the modular equation), then 15) q exp π n), which is the value of q in our definition 1) of G n Since χq) 1/6 {α1 α)q} 1/ [1, p 1], it follows from 11), 1), 15) that 16) G n {α1 α)} 1/ Lastly, we recall that if ϕq) q n, q < 1, then [1, p 10] 17) n ϕq) π K α), when q is given by 1) Ramanujan used modular equations to calculate only a couple of simple invariants in [] This fact the sentence, The values of G n g n are got from the same modular equation [], [, p 5] are the only clues to his methods that Ramanujan provided for us It would seem that if Ramanujan had employed another type of reasoning, he would have dropped some hint about it As mentioned earlier, Watson [31] used modular equations to establish some of Ramanujan s invariants However, for his calculations of G n, it was important that n be a square or a simple multiple of a square We have been able to prove six of the thirteen values for G n, namely, for n 65, 69, 77, 11, 15, 13, by using modular equations As will be seen in our proofs in Section 6, we need some new ideas to effect proofs of these six invariants via modular equations To prove the remaining seven invariants by employing modular equations, we would need modular equations of degrees 31, 1, 3, 53, 79, 9, 101 Apparently, only for degree 31 did Ramanujan derive a modular equation, for he recorded no modular equations for the other six degrees in his notebooks Thus, Ramanujan s methods appear to be even more elusive Watson [30, p ] opined that I believe that fourteen were obtained by Ramanujan by means of the empirical process which I described in the discussion of G 1353 As indicated in our paper [], we believe that Ramanujan found some of these values by the method of [] We are not so confident that Ramanujan used this empirical process, for which Watson offered little explanation In fact, Watson s empirical process is not rigorous However, in Section 7 we shall use class field theory to make Watson s procedure rigorous for a large class of invariants including those 13 invariants examined in this paper, we use the process to calculate two new invariants as well Kronecker s Limit Formula Background Let Qu, v) :y 1 uvz)u v z), where z x iy with y>0the Epstein zeta function ζ Q s) is defined for σ Res>1by 1) ζ Q s): u,v {Qu, v)} s, where the sum is over all pairs of integers u, v) except0,0) It is well known that ζ Q s) can be analytically continued to the entire complex s plane, where ζ Q s) is License or copyright restrictions may apply to redistribution; see

4 1 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG analytic except for a simple pole at s 1The Kronecker limit formula provides the constant term in the Laurent expansion about s 1More precisely, ζ Q s) π s 1 π γ log log y ηz) ) )) Os 1), where γ denotes Euler s constant, ηz) is the Dedekind eta function defined by 3) ηz) :q 1/ q; q) : q 1/ f q), q e πiz,y >0, where the notation f q) is that used by Ramanujan in his notebooks [3] Next, let K be an algebraic number field over the rational numbers Let NA) denote the norm of an ideal A Then the Dedekind zeta function for K is defined by ζ K s) : A NA)) s, σ > 1, where the sum is over all non zero integral ideals A of K Let C K denote the ideal class group of K Then the Dedekind zeta function for an ideal class A of C K is defined by ζs, A) : A ANA)) s, σ > 1 If χ denotes an ideal class character, then the L series for K is given, for σ>1, by ) L K s, χ) : A χa)na)) s A χa)ζs, A), where the former sum is over all non zero integral ideals A of K, the latter sum is over all ideal classes A of C K In the sequel we assume that K is a quadratic field It is well known that [7, p 5] 5) lim s 1)ζ Ks) hκ, s 1 where h is the class number of K, ie, h C K, where π w d, if K is imaginary, 6) κ : logɛ, if K is real d Here w is the number of roots of unity in K, d is the discriminant of K, ɛ is the fundamental unit in K Let ) d L d s) : n s, σ > 1, n n1 where d n) is the Kronecker symbol Then [7, p 5] 7) ζ K s) ζs)l d s), where ζs) denotes the Riemann zeta function License or copyright restrictions may apply to redistribution; see

5 RAMANUJAN S CLASS INVARIANTS 19 Now let d d 1 d, where d 1 > 1, for i 1,,d i 1mod)ord i 0 mod ) Let P denote a prime ideal in K Then a Gauss genus character χ is defined by ) d1, if NP) d 1, NP) χp) ) d, if NP) d 1, NP) ) where di NP) again denotes the Kronecker symbol Note that NP) d if NP) d 1 This definition can be extended to all ideals of K by multiplicativity It is well known that the genus characters form an abelian group, denoted by GK), of order k 1, where k is the number of distinct prime divisors of d Next define G 0 : {A C K : χa) 1,χ GK)}, which is named the principal genus Clearly, G 0 is a subgroup of C K, C K /G 0 is called the genus group Furthermore, C K /G 0 GK) Obviously, A1 A are in the same genus if only if χa 1 )χa )foreachχ GK) Kronecker [7, p 6, Theorem ] proved that, for a genus character χ of K corresponding to the decomposition d d 1 d, ) L K s, χ) L d1 s)l d s) Thus, by ) ), L d1 s)l d s) χa)ζs, A) A C K For a fixed non zero integral ideal B A 1, ζs, A) NB) 9) s NAB)) s NB) s Nλ)) s, σ > 1, A A λ B/U where U is the group of units in K Now assume that K Q m) is an imaginary quadratic field, so m is a squarefree positive integer Recalling that w is the number of roots of unity in K, we see that, from 9), 10) Let Then ζs, A) NB)s w Nλ)) s, σ > 1 λ B λ 0 { m, if m, 3 mod ), Ω 1 m)/, if m 1 mod ) { m, if m, 3 mod ), d m, if m 1 mod ) It is known [15] that each ideal class contains primitive ideals which are Z modules of the form B [a, bω], where a b are rational integers, a>0,a NbΩ), b a/, ais the smallest positive integer in B, NB) a License or copyright restrictions may apply to redistribution; see

6 130 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Let z bω)/a Then, for λ ua vb Ω), 11) Nλ)ua vb Ω))ua vb Ω)) a u vz)u v z) ) 1 a d d u vz)u v z) a Thus, for z bω)/a y Imz d /a), ) 1 d Qu, v) uvz)u v z) a And, from 1), 10), 11), ) s ζs, A) 1 ζ Q s) w d Thus, from ), ) s ζs, A) 1 π w d s 1 πγ π log π log ) d 1) ) s π 1 ) w d loga) log ηz) Os 1) Since, for any nonprincipal genus character χ, χa) 0, A C K it follows from ) 1) that 13) ) s L K s, χ) π χa) 1 ) w log a log ηz) Os 1) d A C K Recall that in the decomposition d d 1 d we assume that d 1 > 1d <0 Let K i Q d i ),i1,by 7), lim s 1)ζ K i s) L di 1), i 1, s 1 Then, by 5) 6), 1) L d1 1) h 1 log ɛ 1 d1 15) L d 1) h π w d, where h i is the class number of K i,i 1,,ɛ 1 is the fundamental unit of K 1, w is the number of roots of unity in K Thus, setting s 1 in 13) using License or copyright restrictions may apply to redistribution; see

7 RAMANUJAN S CLASS INVARIANTS 131 ), we deduce that L d1 1)L d 1) π w χa) 1 ) 16) d log a log ηz) A C K Thus, setting 17) F A) ηz) / a, where z bω)/a, with [a, b Ω] A 1, we conclude from 1) 17) that, for χ nonprincipal [7, p 7], wh 1 h log ɛ 1 χa)logfa), w A C K or 1) ɛ wh1h/w 1 F A) χa) A C K We remark that 1) was utilized by K G Ramanathan [17], [1], [19], [0], [1] to calculate class invariants, values of the Rogers Ramanujan continued fraction, certain other invariants of Ramanujan 3 Two Primary Theorems Let τ m Then, by 11) 1), it is easily seen that ) τ 1 η 31) ητ) 1/ G m Equalities 1) 31) are the key ingredients for deriving formulas that will enable us to calculate G m In this section, we consider two different genus structures, the two theorems that we prove can be utilized to determine G m for m 65, 69, 77, 11, 15, 05, 13, 65, 301, 5, 505 For m 17, 553, the genus structure is of a third type, this type will be examined in Section 5 In each case, K Q m) has class number, the number of genera equals Thus, each genus contains exactly two ideal classes Also note that A A 1 are clearly in the same genus Throughout this paper, for simplicity, we use the notation for a primitive ideal to denote the ideal class containing it; this abuse of notation should not cause difficulty Theorem 31 Let m 1mod),where m is a positive squarefree integer with prime divisor p Let K Q m) be an imaginary quadratic field such that each genus contains exactly two ideal classes such that the principal genus G 0 contains the classes [1, Ω] [p, p Ω] Let G 1 be a nonprincipal genus containing the two classes [, 1Ω] [p, Ω] Then Gm G m/p ) h/ χg 1) 1 ɛ wh1h/w 1, where h, h 1, h are the class numbers of K, Q d 1 ), Q d ), respectively, w w are the numbers of roots of unity in K Q d ), respectively, ɛ 1 is the License or copyright restrictions may apply to redistribution; see

8 13 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG fundamental unit in Q d 1 ), the product is over all characters χ with χg 1 ) 1), associated with the decomposition d d 1 d, therefore d 1,d,h 1,h,w, ɛ 1 are dependent on χ Proof Each of the ideals [1, Ω], [p, p Ω],[, 1Ω], [p, Ω] is ambiguous If A A is any one of these ideals, then A A 1,AA 1, A A 1 For any ideal class B/ G 0 G 1,it is not difficult to see that [0, p 77] χb) 0, which implies that χg 1) 1 χg 1) 1 F B) χb) 1, where F B) is defined by 17) Therefore, by 1), 3) ɛ wh1h/w 1 F A) χa) F A) χa)h/, A G 0 G 1 A G 0 G 1 χg 1) 1 χg 1) 1 since the number of genus characters equals h/, so the number of genus characters with χg 1 ) 1ish/ Let A 0 [1,Ω],A 0 [p, p Ω],A 1 [,1Ω], A 1 [p, Ω] Then, by 3), 33) χg 1) 1 ɛ wh1h/w 1 F A1 )/F A 0 ) F A 0 )/F A 1 ) ) h/ By 17) 31), F A 1 ) F A 0 ) η Ω1 )/ 3) η G Ω) m Let Ω Ω/p m/p Again, by 17) 31), F A 0 ) F A 1 ) η Ωp p )/ p η Ω p ) η Ω 1 )/ 35) η Ω G m/p ) The theorem now follows from 33) 35) Theorem 3 Let m 1mod),where m is a positive squarefree integer with prime divisor p Let K Q m) be an imaginary quadratic field such that each genus contains exactly two ideal classes such that the principal genus G 0 contains the classes [1, Ω] [p, Ω] Let G 1 be a nonprincipal genus containing the two classes [, 1Ω] [p, p Ω] Then h/ Gm G m/p ) ɛ wh1h/w 1, χg 1) 1 where h, h 1, h are the class numbers of K, Q d 1 ), Q d ), respectively, w w are the numbers of roots of unity in K Q d ), respectively, ɛ 1 is the fundamental unit in Q d 1 ), the product is over all characters χ with χg 1 ) 1), associated with the decomposition d d 1 d, therefore d 1,d,h 1,h,w, ɛ 1 are dependent on χ License or copyright restrictions may apply to redistribution; see

9 RAMANUJAN S CLASS INVARIANTS 133 The proof of Theorem 3 is analogous to that for Theorem 31, so we omit it We say that m is of the first kind or second kind according as it satisfies the conditions of Theorem 31 or Theorem 3, respectively It is not difficult to show that [1, Ω], [, 1Ω],[p, Ω], [p, p Ω] are representatives of different ideal classes [15] Before commencing our calculations we need three modular equations of Ramanujan [1, pp 31,, 315] Lemma 33 Modular Equation of Degree 3) Let P {16αβ1 α)1 β)} 1/ Q ) 1/ β1 β) α1 α) Then Q 1 Q P 1 ) 0 P Lemma 3 Modular Equation of Degree 5) Let P {16αβ1 α)1 β)} 1/1 Q ) 1/ β1 β) α1 α) Then Q 1 Q P 1 ) 0 P Lemma 35 Modular Equation of Degree 7) Let P {16αβ1 α)1 β)} 1/ Q ) 1/6 β1 β) α1 α) Then Q 1 Q 7 P 1 ) P Let q exp π/ n) Since G n G 1/n [], [, p 3], by 16), G n {α1 α)} 1/ If β has degree p over α, then G n/p G p /n {β1 β)} 1/ In summary, we can express the equalities of Lemmas in terms of G n G n/p,p3,5,7,respectively, by employing the formulas 36) G n {α1 α)} 1/ G n/p {β1 β)} 1/ The class numbers cited below for d < 500 can be found in tables in the texts of Borevich Shafarevich [6, pp 6], H Cohen [10, pp ], for 0 <d<10, 000 in the book of DA Buell [, pp 3] Lists of fundamental units can be found in [6] for d 101), the book by M Pohst H Zassenhaus [16, pp 3 35] up to d 99), the tables of R Kortum G McNiel [1] up to d 10,000) In Cohen s book [10, pp 6 7], there is a table providing the ideal class structure for Q d),d 97 for Q d),d 97 License or copyright restrictions may apply to redistribution; see

10 13 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Calculations of 11 Class Invariants Theorem 1 ) 1/ ) 1/ G Proof The following table summarizes the needed information about ideal classes their characters 1/ d 1 d χ G C [1, Ω] 1 60 χ 0 G 0 [10, 5Ω] [, 1Ω] 5 5 χ 1 G 1 [5, Ω] [3, 1Ω] 13 0 χ G [3, 1Ω] [6, 1Ω] 65 χ 3 G 3 [6, 1Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Note that 65 is of the first kind Applying Theorem 31 with h w, we find that 1) G65 G 13/5 ) ) ) 53 3 Let Q G 65 /G 13/5 ) 3 P G 65 G 13/5 ) Then, by 1), ) ) 3/ ) 3/ ) 1/ ) 1/ Q By Lemma 3, 3) Now, by ), ), by ), 5) P 1 Q Q 1 ) QQ 1 ) 16 Q Q 1 ) 16Q Q 1 5)5 13 1) 5 )5 13 1) ), Thus, by 3) ), 6) P 1 1 Q Q ) License or copyright restrictions may apply to redistribution; see

11 RAMANUJAN S CLASS INVARIANTS 135 Thus, by ) 6), G 65 Q 1/6 P 1/ ) 1/ ) 1/1 533 Thus, it remains to show that ) )) 1/ 9 65 which is easily shown by a routine calculation Theorem 5 ) 1/ ) 1/ 3 G , Proof We summarize the needed information in the following table 3 3 1/ d 1 d χ G C [1, Ω] 1 76 χ 0 G 0 [6, 3Ω] [, 1Ω] 9 3 χ 1 G 1 [3, Ω] [5, 1Ω] 69 χ G [5, 1Ω] [7, 1Ω] 1 3 χ 3 G 3 [7, 1Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ We apply Theorem 31 with h w,as 69 is of the first kind Thus, ) G69 5 3) 1/3 53 ) 1/ 69 7) G 3/3 Let Q G 69 /G 3/3 ) 6 P G 69 G 3/3 ) 3 By 7), Q 5 3) 1/ 5 3 ) 3/ 69 ) 5 ) ) 1/ 5 ) ) 3/ 3 By Lemma 33, 9) From ), 10) P 1 1 Q Q 1 ) 1 QQ 1 ) 3 Q Q 1 Q Q ), License or copyright restrictions may apply to redistribution; see

12 136 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG, from 10), Q Q 1 ) ) Putting these calculations in 9), we find that 11) P ) By ), G 69 Q 1/1 P 1/6 5 ) 1/ ) 1/ 3 P 1/6, thus, by 11), it remains to show that ) This can be achieved by a straightforward computation Theorem 3 G 77 3 ) 1/ ) 1/ Proof We compose the following table giving needed information about ideal classes characters 1/ d 1 d χ G C [1, Ω] 1 30 χ 0 G 0 [1, 7Ω] [, 1Ω] 11 χ 1 G 1 [7, Ω] [3, 1Ω] 77 χ G [3, 1Ω] [6, 1Ω] 7 χ 3 G 3 [6, 1Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ We see from the table that 77 is of the first kind Thus, by Theorem 31, since h w, ) G77 Q: 3 9 ) 1/ 77 7) 3 ) ) 7) G 11/7 If P G 77 G 11/7 ) 3, then, from Lemma 33, 13) Now P 1 Q Q 1 7 QQ 1 7) 3 Q Q ) License or copyright restrictions may apply to redistribution; see

13 RAMANUJAN S CLASS INVARIANTS 137 Using this in 13), we find that 1) By 1), P ) 11 G 77 Q 1/ P 1/6 3 ) 1/ ) 1/ P 1/6, thus by 1) it remains to show that ) 11, which is readily shown by a straightforward calculation Theorem G ) 1/ 7 ) 1/ Proof We record the necessary information in the following table / d 1 d χ G C [1, Ω] 1 56 χ 0 G 0 [6, 3Ω] [, 1Ω] 1 3 χ 1 G 1 [3, Ω] [5, Ω] 11 χ G [5, Ω] [10, 3Ω] 1 7 χ 3 G 3 [10, 3Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ We see that 11 is again of the first kind Applying Theorem 31, we find that, since h w, ) G11 15) 7 7) 1/ ) 1/ G 7/3 Let Q G 11 /G 7/3 ) 6 Then, by 15), Q 7 7) 1/ 95 11) 3/ 16) Let P G 11 G 7/3 ) 3 Then, by Lemma 33, 7 7 ) 3 7) 3/ 7 7) 1/ ) 1/ 17) P 1 Q Q 1 ) QQ 1 ) 3 License or copyright restrictions may apply to redistribution; see

14 13 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG From the last representation of Q in 16), Q Q )), so Q Q 1 Q Q Using these calculations in 17), we deduce that P 1 1 1) ) Hence, by 16) 1), G 11 Q 1/1 P 1/6 It thus remains to show that ) 1/ ) 1/ /6 3)) which is a straightforward, albeit laborious, task Theorem 5 G 15 ) 1/ 95 5) 1/ Proof We compose the following table ), / d 1 d χ G C [1, Ω] 1 50 χ 0 G 0 [5, Ω] [, 1Ω] 9 0 χ 1 G 1 [10, 5Ω] [7, 3Ω] χ G [7, 3Ω] [11, 3Ω] 15 χ 3 G 3 [11, 3Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Thus, 15 is of the second kind Thus, by Theorem 3, since h w, ) ) 6 ) G 15 G 9/5 ) 5) Hence, ) ) P 1 : G 15 G 9/5 ) 5) License or copyright restrictions may apply to redistribution; see

15 RAMANUJAN S CLASS INVARIANTS 139 By Lemma 3, with Q G 15 /G 9/5 ) 3, we have 0) Q P 1 P P 1 P ) 1 By 19), we readliy find that P 1 P 95 5, so, by 0), Q 9 5 1) Thus, by 19) 1), ) 1/ 9 5 G 15 P 1/ Q 1/6 5) 1/ /6 15) Hence, it remains to show that which is readily shown Theorem 6 ) ) 1/ 1 G Proof We record the following table , 1 1 d 1 d χ G C [1, Ω] 1 0 χ 0 G 0 [5, Ω] [, 1Ω] 5 16 χ 1 G 1 [10, 5Ω] [11, Ω] 05 χ G [11, Ω] [13, Ω] 1 0 χ 3 G 3 [13, Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Note that 05 is of the second kind Applying Theorem 3 with h w,we deduce that ) 51 P : G 05 G 1/5 ) 33 ) 05 ) 73 ) ) 1 License or copyright restrictions may apply to redistribution; see

16 10 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Letting Q G 05 /G 1/5 ) 3, we deduce from Lemma 3 that 3) Q P 1 P) P 1 P) 1 From ), Thus, from 3), ) Q 1 P 1 P ) 1 If follows from ) ) that ) 51 G 05 P 1/ Q 1/6 3 5 ) 1/ /6 1)) It thus remains to show that ) This is more readily accomplished if we first note that Theorem 7 G ) 1/ 71 Proof We have the following table 59 7 ) 1/ / d 1 d χ G C [1, Ω] 1 5 χ 0 G 0 [6, 3Ω] [, 1Ω] 3 χ 1 G 1 [3, Ω] [7, Ω] 13 χ G [7, Ω] [1, 5Ω] 1 71 χ 3 G 3 [1, 5Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ License or copyright restrictions may apply to redistribution; see

17 RAMANUJAN S CLASS INVARIANTS 11 Observe that 13 is of the first kind Applying Theorem 31 with h w,we find that ) Q /3 G13 : ) 1/ ) 1/ 13 G 71/ ) / ) 71, so that 5) 597 ) ) 3/ ) 71 Q ) 1/ Let P G 13 G 71/3 ) 3 Then, by Lemma 33, 6) P 1 1 By 5) moderate calculations, Q Q 1 ) QQ 1 ) 3 Q Q 1 ) ) 3) 1 Q Q 1 ) 1 1 Q Q ) Thus, by 6), 7) P 1 ) ) ) Thus, by 5) 7), G 13 Q 1/1 P 1/6 5 3 ) 1/ ) 1/ ) ) 1/6 3) Hence, it remains to show that ) ) which is accomplished by a direct calculation Theorem G 65 ) 1/ ) 1/ Proof The following table is easily verified , / License or copyright restrictions may apply to redistribution; see

18 1 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG d 1 d χ G C [1, Ω] 060 χ 0 G 0 [10, 5Ω] [, 1Ω] 5 1 χ 1 G 1 [5, Ω] [7, 1Ω] 53 0 χ G [7, 1Ω] [1, 1Ω] 65 χ 3 G 3 [1, 1Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Note that 65 is of the first kind Applying Theorem 31 with h w, we find that ) ) 6 ) Q /3 G :, so that G 53/5 ) ) 9/ ) 3/ ) 1/ ) 1/ Q Let P G 65 G 53/5 ) Then, by Lemma 3, P 1 Q Q 1 ) QQ 1 ) 16 9) By using ) the identity Q Q 1 Q Q in 9), we find that P ) ) By ) 30), ) 1/ G 65 Q 1/6 P 1/ ) 1/ / 65)) 55 Hence, it remains to show that ) , which is easy to establish Theorem 9 G ) 1/ ) 1/ Proof We compose the following table 7 3 1/ License or copyright restrictions may apply to redistribution; see

19 RAMANUJAN S CLASS INVARIANTS 13 d 1 d χ G C [1, Ω] 0 χ 0 G 0 [1, 7Ω] [, 1Ω] 3 χ 1 G 1 [7, Ω] [5, Ω] 301 χ G [5, Ω] 17 7 χ 3 G 3 [10, 3Ω] [10, 3Ω] χg 0) χg 1) χg ) χg 3) h1 h w ɛ Thus, 301 is of the first kind Applying Theorem 31 with h w,we find that ) G301 Q : ) 1/ 301 7) G 3/ ) 7 31) 7) Let P G 301 G 3/7 ) 3 Then, by Lemma 35 31), 3) P 1 1 Q Q 1 7) 1 QQ 1 7) ) ) Therefore, by 31) 3), ) 1/ G 301 Q 1/ P 1/6 3 7) 1/ ) /6 3)) It remains to show that ) ) which is a routine task Theorem G 5 5) 1/ 5 1 Proof We form the following table 7 3 3, ) 1/ License or copyright restrictions may apply to redistribution; see

20 1 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG d 1 d χ G C [1, Ω] 70 χ 0 G 0 [5, Ω] [, 1Ω] χ 1 G 1 [10, 5Ω] [13, 6Ω] 5 χ G [13, 6Ω] [19, 7Ω] 9 0 χ 3 G 3 [19, 7Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Thus, 5 is of the second kind Applying Theorem 3 with h w, we deduce that ) 1 ) P : G 5 G 9/5 ), so that 33) P 1 9 ) 5 1 5) Let Q G 5 /G 9/5 ) 3 Then, by Lemma 3 33), 3) Q P 1 P) P 1 P) Therefore, by 33) 3), ) 1/ G 5 P 1/ Q 1/ ) 1/ /6 9) It thus remains to show that By first squaring the binomial on the right side then cubing the resulting expression, we can easily verify the desired equality Theorem 11 G 505 ) 1/ 51 5) 1/ ) 1/ / 505 Proof We compose the following table License or copyright restrictions may apply to redistribution; see

21 RAMANUJAN S CLASS INVARIANTS 15 d 1 d χ G C [1, Ω] 1 00 χ 0 G 0 [5, Ω] [, 1Ω] 5 0 χ 1 G 1 [10, 5Ω] [11, 1Ω] χ G [11, 1Ω] [, 1Ω] 505 χ 3 G 3 [, 1Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ Hence, 505 is of the second kind Applying Theorem 3 with h w, we find that ) 1 51 P : G 505 G 101/5 ) ), so that ) 7 51 P ) ) 51 5) ) 73 ) 5 35) 5) ) Let Q G 505 /G 101/5 ) 3 Then, by Lemma 3 35), 36) Q P 1 P) P 1 P) ) Therefore, by 35) 36), G 505 P 1/ Q 1/6 ) 1/ 51 5) 1/ ) 1/ ) /6 505) Thus, it remains to show that ) , which is straightforward 5 G 17 G 553 The genus structures for Q 17) Q 553) are different from those for the eleven imaginary quadratic fields addressed in Section, so G 17 G 553 must be calculated by another means This we accomplish in this section License or copyright restrictions may apply to redistribution; see

22 16 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Lemma 51 Let m denote a positive integer with 7 m Let τ m/7 Q G m /G m/9 ) Then 51) ητ)η τ 1 ) η7τ)η 7τ 1 )) 9 ητ)η τ1 ) )) η7τ)η 7τ 1 Q 3/ Q 1/ Q 1/ Q 3/ Proof With q exp π m/7), it follows from 3) that ητ)η τ 1 ) 5) η7τ)η 7τ 1 )) f q )f q) q 3/ f q 1 )f q 7 ) Next, by an entry from Ramanujan s second notebook [, p 09, Entry 55], f q)f q ) 9q3/ f q 7 )f q 1 ) q 3/ f q 7 )f q 1 ) f q)f q f 6 q )f 6 q 7 ) ) q 3/ f 6 q)f 6 q 1 ) f q )f q 7 ) q 1/ f q)f q 1 ) q1/ f q)f q 1 ) f q )f q 7 ) q3/ f 6 q)f 6 q 1 ) f 6 q )f 6 q 7 ) Multiplying both sides by q 3/ thenreplacingqby q, we find that f q)f q ) f q 7 )f q 1 ) f q 7 )f q 1 ) 9q3 f q)f q f 6 q )f 6 q 7 ) ) f 6 q)f 6 q 1 ) qf q )f q 7 ) f q)f q 1 ) q f q)f q 1 ) f q )f q 7 q3 f 6 q)f 6 q 1 53) ) ) f 6 q )f 6 q 7 ) Recall that q exp π m/7) recall that G m/9 is then given by 1) Thus, G m 1/ q 7/ q 7 ; q 1 ) Hence, 5) Gm G m/9 ) q7 ; q 1 ) q 1/ q; q ) q ; q ) q 7 ; q 7 ) q 1/ q; q) q1 ; q 1 ) f q )f q 7 ) q 1/ f q)f q 1 ) Dividing 53) by q 3/ substituting 5) 5) into the resulting equality, we deduce 51) to complete the proof Theorem 5 Let m be a squarefree positive integer with 7 m m 1mod) Let K Q m) be an imaginary quadratic field such that each genus contains exactly two classes such that the principal genus G 0 comprises [1, Ω] [, 1 Ω], while [7, Ω] [1, 7Ω] form a nonprincipal genus G 1 Then { 1 ητ)η τ 1 ) h/ 7 η7τ)η 55) ))} 7τ 1 ɛ wh1h/w 1, χg 1) 1 where h, h 1, h are the class numbers of K, Q d 1 ), Q d ), respectively, w w are the numbers of roots of unity in K Q d ), respectively, ɛ 1 is the fundamental unit in Q d 1 ), the product is over all characters χ with χg 1 ) 1), associated with the decomposition d d 1 d, therefore d 1,d,h 1,h,w, ɛ 1 are dependent on χ License or copyright restrictions may apply to redistribution; see

23 RAMANUJAN S CLASS INVARIANTS 17 Proof Let A 0 [1,Ω],A 0 [,1Ω],A 1 [7,Ω], A 1 [1,7Ω] Then by the same reasoning that we used in the proof of Theorem 31, F A1 )F A ) h/ 56) 1) F A 0 )F A 0 ) ɛ wh1h/w 1 By 17), F A 0 )η Ω) η 7τ), Ω1 F A 0 )η FA 1 )η Ω 7 χg 1) 1 ) / η 7τ1 ) / 7η τ)/ 7, ) /, ) Ω7 F A 1 )η / ) τ 1 1 η / 1 1 Substituting these values into 56) recalling that the number of genus characters χ with χg 1 ) 1isequaltoh/, we deduce 55) to complete the proof Theorem 53 G / 15 7 Proof We set up a table to summarize some information that we need 1/ d 1 d χ G C [1, Ω] 1 6 χ 0 G 0 [, 1Ω] [7, Ω] 1 7 χ 1 G 1 [1, 7Ω] [11, 5Ω] 17 χ G [11, 5 Ω] [13, Ω] 31 χ 3 G 3 [13, Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ It is clear that Q 17) satisfies the conditions of Theorem 5 Thus, since h w,we deduce that 1 ητ)η τ 1 ) )) 7 η7τ)η 7τ ) ) 1/ so that where 57) ) ), ητ)η τ1 ) )) η7τ)η 7τ 1 7ɛ, ɛ ) 1/ ) 1/ License or copyright restrictions may apply to redistribution; see

24 1 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG It follows from 51) that 5) Q 3/ Q 1/ Q 1/ Q 3/ 7ɛ ɛ 1 ), where Q G 17 /G 31/7 ) By an elementary calculation, ɛ ɛ 1 ) ɛ ɛ ) 710 7) ) Let x Q 1/ Q 1/ Then 5) can be recast in the form x 3 11x ) ) It is now obvious that 59) x Solving 59) for Q 1/, we find that 510) Q 1/ ) ) 7) Now let P G 17 G 31/7 ) 3 Using Lemma 35 59), we deduce that P P 1 1 Q Q 1 7 ) 1 x 9) ) Solving for P 1, we find that 511) P Thus, from 510) 511), G 17 P 1/6 Q 1/ 11 7 which completes the proof / 1/, License or copyright restrictions may apply to redistribution; see

25 RAMANUJAN S CLASS INVARIANTS 19 Theorem 5 G Proof We set up the following table to summarize the information that we need 1/ 1/ d 1 d χ G C [1, Ω] χ 0 G 0 [, 1Ω] [7, Ω] 79 χ 1 G 1 [1, 7Ω] [17, 5Ω] 553 χ G [17, 5 Ω] [19, 6Ω] χ 3 G 3 [19, 6 Ω] χg 0) χg 1) χg ) χg 3) h 1 h w ɛ , 635, 37, 07 6, 56, 17, It is clear that Q 553) satisfies the hypotheses of Theorem 5 Thus, since h w, 1 7 ητ)η τ1 ) so that 51) where 513) η7τ)η 7τ 1 )) 3 7) 5 6, 635, 37, 07 6, 56, 17, ) 1/, ητ)η τ1 ) )) η7τ)η 7τ 1 7ɛ, ɛ 51, 0 19, 307 7) 1/ 11, 7 76,76 79) 1/ Then an elementary calculation gives 51) ɛ ɛ 1 ) ɛ ɛ 13, 650, 096, 11 16, 161, 9, 5 79) ) ) By Lemma 51 51) 51), with Q G 553 /G 79/7 ), Q 3/ Q 1/ Q 1/ Q 3/ 7ɛ ɛ 1 ) ) If x Q 1/ Q 1/, then the foregoing equality may be written in the form x 3 11x ) 79) 11, License or copyright restrictions may apply to redistribution; see

26 150 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG from which it is obvious that 515) x Q 1/ Q 1/ Solving for Q 1/, we readily find that Q 1/ ) 79) ) Now let P G 553 G 79/7 ) 3 Then, by Lemma ), P P 1 )QQ 1 7x Solving for P 1, we find that P , 91, 1 5, 165, 776 ) ) Thus, by 516) 517), G 553 Q 1/ P 1/6 the proof is complete Class Invariants Modular Equations In this section we establish six of Ramanujan s class invariants by using tools well known to Ramanujan, in particular, modular equations Second Proof of Theorem 1 From 11) 3) it is easily seen that 61) f q) f q ) χ q) Using this equality, we rewrite two of Ramanujan s eta function identities in terms of χ Thus [, pp 06, 11] f q)f q ) 13q3/ f q 13 )f q 6 ) 3 ) q 3/ f q 13 )f q 6 ) f q)f q q 1/ χ q13 ) ) χ q) ) q 1/ χ q13 ) q 1/ χ q) ) χ q) χ q 13 q 1/ χ q) ) 3 6) ) χ q 13 ) 1/ 1/, License or copyright restrictions may apply to redistribution; see

27 RAMANUJAN S CLASS INVARIANTS ) f q)f q ) 5q1/ f q 5 )f q 10 ) q 1/ f q 5 )f q 10 ) f q)f q ) 6) ) 3 ) 3 q 1/6 χ q5 ) 1/6 χ q) q χ q) χ q 5 ) Replace q by q in 6) then set q exp π 5/13) If A : e 3π/) 5/13 fe π 5/13 )f e π 5/13 ) fe π 65 )f e π 65 ) B : e π/) 65 5/13 χe π ) 65) χe π, 5/13 ) then 6) can be recast in the form 66) A 13A 1 B 3 B B 1 B 3 Next, replace q by q in 63) then set q exp π 13/5) If 67) A : e π/) 13/5 fe π 13/5 )f e π 13/5 ) fe π 65 )f e π 65 ) 6) B : e π/6) 65 13/5 χe π ) χe π, 13/5 ) then 63) takes the shape 69) A 5A 1 B 3 B 3 We shall prove that ) B B A 5 A Now, by 1), G n 1/ e πn/ χe πn ) G 1/n 1/ e π/n) χe π/n ) Since G n G 1/n, we find that χe πn )e π/)1/ n 611) n) χe π/n ) This could also be proved by using 61) along with the transformation formula for f) In particular, if n 5/13, 611) yields the equality 61) χe π5/13 )e π/365) χe π13/5 ) The aforementioned transformation formula for f q) is given by [1, p 3, Entry 7iii)] 613) e a/1 a 1/ f e a )e b/1 b 1/ f e b ), where a, b > 0 with ab π If a π 5/13, so that b π 13/5, then we deduce from 613) that 61) f e π 5/13 )13/5) 1/ e π/3 65) f e π 13/5 ) License or copyright restrictions may apply to redistribution; see

28 15 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG First, from 65) 61), B eπ/)5/13 χe π 65 ) e π/3 65) χe π13/5 ) e π/6) 65 13/5 χe π ) χe π B, 13/5 ) by 6) Thus, the first equality of 610) has been demonstrated Second, by 6), 61) with q exp π 5/13), 61), 61), lastly 61) with q exp π 13/5), A e 3π/) 5/13 χe π5/13 )f e π5/13 ) fe π 65 )f e π 65 ) 13 13/5 5 eπ/) 13/5 χe π )f e π13/5 ) fe π 65 )f e π 65 ) 13 13/5 5 eπ/) 13/5 fe π )f e π13/5 ) fe π 65 )f e π 65 ) 13 5 A, by 67) Thus, the second equality of 610) has been established Employing 610) in 69), we find that 5 13 A 65A 1 B 3 B 3 B B 1 ) 3 3B B 1 ) Dividing both sides by u : B B 1 0),we find that u 7)u 3 Solving for u, we find that u 65 1)/ Thus, since clearly B>1, B B 1 Now solving for B, we find that ) B, where in solving the quadratic equation we took the plus sign since B>0 If q exp π 13/5), then q 5 exp π 65) Hence, from 1) 65), we readily see that B G 65 /G 13/5 Furthermore, from 16), G 13/5 {α1 α)} 1/ Hence, if β has degree 5 over α, then G 65 {β1 β)} 1/ We now employ Lemma 3, where it is to be noted that P G 65 G 13/5 ) Q B 3 G 65 /G 13/5 ) 3 We already know Q from 615) To determine P from Lemma 3, we first calculate Q Q 1 B 3 B 3 BB 1 ){BB 1 ) 3} 657 ) ) License or copyright restrictions may apply to redistribution; see

29 RAMANUJAN S CLASS INVARIANTS 153 Thus, using 616) in Lemma 3 solving for P 1, we find that P ) 617) 65, since P>0 Hence, by 615) 617), / G 65 B 1/ P 1/ / 61) 65)) We must show that 61) can be transformed into the form of Theorem 1 First, ) Second, 60) ) 1/ )1 65) ) ) 13 ) ) Putting 619) 60) in 61), we complete the proof ) 1/ Before commencing our second proof of Theorem, we establish a general principle Let p r denote coprime, positive integers Set q exp π p/r) q exp π pr), let β have degree r over α Then, by 16), 61) G p/r {α1 α)} 1/ G pr {β1 β)} 1/ Furthermore, from 1) 15), 6) K 1 α) K α) p r, from the defintion 13) of a modular equation, 63) r K 1 α) K α) K 1 β) K β) If we solve 6) for r substitute this in 63), we find that p K α) K 1 α) K 1 β) K β) License or copyright restrictions may apply to redistribution; see

30 15 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG From the last equality we conclude: 6) If β has degree r over α, then β has degree p over 1 α Furthermore, from 17) 6), 65) ϕ e π p/r ) ϕ e π r/p ) K α) r K 1 α) p Second Proof of Theorem We need two of Ramanujan s modular equations of degree 3 [1, p 11, Entry 15i), ii)] If β has degree 3 over α, then 66) 67) αβ) 1/ {1 α)1 β)} 1/ /3 {αβ1 α)1 β)} 1/ 1 1αβ) 1/ {1 α)1 β)} 1/ /3 {αβ1 α)1 β)} 1/1 { 1αβ) 1/ {1 α)1 β)} 1/)} 1/ We also need two of Ramanujan s modular equations of degree 3 The first is given by Lemma 33, while the second is given by [1, p 31, Entry 5ix)] 6) {α1 β)} 1/ {1 α)β} 1/ {αβ1 α)1 β)} 1/ We shall apply 6) with r 3p3Thus, β has degree 3 over 1 α) Thus, replacing α by 1 α), from 66) 67), we find that, respectively, 69) 630) {1 α)β} 1/ {α1 β)} 1/ /3 {αβ1 α)1 β)} 1/ 1 1{1 α)β} 1/ {α1 β)} 1/ /3 {αβ1 α)1 β)} 1/1 { 1{1 α)β} 1/ {α1 β)} 1/)} 1/ For brevity, in the remainder of the proof, set G G 69 G G 3/3 By 61), we can rewrite 69) in the form {1 α)β} 1/ {α1 β)} 1/ 1 GG ) 1 Setting u GG ) 1 squaring both sides, we deduce that 631) {1 α)β} 1/ {α1 β)} 1/ 1u u u 3 Substituting 631) into 630), we find that u u u 3 1{1 α)β} 1/ {α1 β)} 1/) 1/ 63) Then, using 6) in 63), we deduce that u u u 3 1 u 3 ) 1/ Squaring both sides simplifying, we arrive at u u u 3 u u 5 u 6 0, License or copyright restrictions may apply to redistribution; see

31 RAMANUJAN S CLASS INVARIANTS 155 which, with x u 1/u, is equivalent to u 3 u 3 ) u u ) u u 1 ) x 3 3x) x ) x Simplifying, we find that x 3 x 9x 3 0 By inspection, we verify that is a root Now G n is a monotonically increasing function of n, it is not difficult to numerically check that the root that we seek is greater than Thus, x 3 x 30, so x 3 3)/ Since x u 1/u, we find that 1 u ), since u<1 We now apply Lemma 33 Noting that P u 3, we see that we want to calculate u 3 u 3 u 3 u 3 ) x 3 3x) Thus, by Lemma 33, ) G 6 G Solving for G/G, we deduce that 63) G G 3 ) ) 3 ) 6 G G u 3 u 3 ) ) 1/6 Thus, by 633) 63), G G G u 1/ ) 1/ To complete the proof, it suffices to show that ) which is a straightforward task 3 3 1/ 5 ) ) 3 3, License or copyright restrictions may apply to redistribution; see

32 156 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Second Proof of Theorem 3 We need two of Ramanujan s modular equations of both degrees 7 11 If β has degree 7 over α, then [1, pp 31, 315, Entry 19i), viii)] 635) 636) αβ) 1/ {1 α)1 β)} 1/ 1 m 7 m αβ) 1/ {1 α)1 β)} 1/) αβ) 1/ {1 α)1 β)} 1/), where m ϕ q)/ϕ q 7 ) If β has degree 11 over α, then [1, p 363, Entry 7i), ii)] 637) 63) αβ) 1/ {1 α)1 β)} 1/ {16αβ1 α)1 β)} 1/1 1 m 11 m αβ) 1/ {1 α)1 β)} 1/) αβ) 1/ {1 α)1 β)} 1/), where m ϕ q)/ϕ q 11 ) If q exp π 11/7), by 61), G 11/7 {α1 α)} 1/ G 77 {β1 β)} 1/ Thus, setting u G 77 G 11/7 ) 1, we deduce from 635) that αβ) 1/ {1 α)1 β)} 1/) αβ) 1/ {1 α)1 β)} 1/) {α1 α)β1 β)} 1/ 1 u 3 αβ) 1/ {1 α)1 β)} 1/ αβ) 1/ {1 α)1 β)} 1/) Thus, from 636), 639) {α1 α)β1 β)} 1/ 1 u 3 m 7 m 1 u 3) 1/ 3 u 3 ), where m ϕ e π11/7 )/ϕ e π77 ) Let q exp π 7/11), note that u G 77 G 11/7 ) 1 G 77 G 7/11 ) 1 Thus, by 637), αβ) 1/ {1 α)1 β)} 1/ 1 u αβ) 1/ {1 α)1 β)} 1/ ) αβ) 1/ {1 α)1 β)} 1/) {α1 α)β1 β)} 1/ 1 u ) u 6 License or copyright restrictions may apply to redistribution; see

33 RAMANUJAN S CLASS INVARIANTS 157 Hence, from 63), 60) m 11 m 1 u ) u 6) 1/ 5 u ), where m ϕ e π 7/11 )/ϕ e π 77 ) From 65), we see that Since m m m 11 m 7 we deduce from 639) 60) that 1 u 3) 1/ 3 u 3 ) 11 7 m 7 m Squaring both sides simplifying, we find that 7 m 11 ) 11 m, u ) u 6) 1/ 5 u ) u u 9 9u 37u 6 3u 66 u 3 10u 19 0 Isolating the terms involving on one side of the equation, squaring both sides, simplifying, factoring, we deduce that 61) u u 6 7u u 1) 196u 1 11u 10 60u 136u u 509u 361) 0 Now x : u is an algebraic integer see Lemma 7) so must be a root of a monic irreducible polynomial The latter polynomial in 61) is irreducible, so x must be a root of the former polynomial in 61) Alternatively, we used Mathematica to numerically determine the roots of the latter polynomial found that u is not one of these roots Thus, x x 3 7x x1x x1/x) x 1/x)5 ) 0 Since x 1/x > 1, x 1 x 11 Thus, u 1 u x 1 6 x 11 Since u<1,we find that 6) 1 u License or copyright restrictions may apply to redistribution; see

34 15 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Lastly, we apply Lemma 35 Since P u 3, we deduce, by 6), that Hence, 63) Q Q 1 u 3 u 3 ) 7 uu 1 ) 3 3u u 1 ) ) ) )1 11) Q ) ) In conclusion, by 6) 63), G 77 Q 1/ u 1/ 3 7) 1/ 11 7 ) 1/ ) the proof is complete Second Proof of Theorem We need two of Ramanujan s modular equations, one of degree 3 one of degree 7 If β has degree 3 over α [1, p 31, Entry 5ix)], 6) {α1 β)} 1/ {β1 α)} 1/ {αβ1 α)1 β)} 1/ If β is of degree 7 over α [1, p, Entry 3i)], 65) 1 1αβ) 1/ {1 α)1 β)} 1/)) 1/ 1αβ) 1/ {1 α)1 β)} 1/ 1/3 {αβ1 α)1 β)} 1/ 1αβ) 1/ {1 α)1 β)} 1/) Let q exp π 7/3) Then, by 61), G : G 7/3 {α1 α)} 1/ G : G 11 {β1 β)} 1/ Applying 6) with r 3p7,we find that β has degree 7 over 1 α) when β has degree 3 over α Thus, by 65), 1 1{1 α)β} 1/ {α1 β)} 1/)) 1/ 66) 1 {1 α)β} 1/ {α1 β)} 1/ 1/3 {α1 α)β1 β)} 1/ 1{1 α)β} 1/ {α1 β)} 1/) 1/, License or copyright restrictions may apply to redistribution; see

35 RAMANUJAN S CLASS INVARIANTS 159 If u : GG ) 1, by 6), 67) {α1 β)} 1/ {β1 α)} 1/ u 3 Hence, {α1 β)} 1/ {β1 α)} 1/) 6) {α1 β)} 1/ {β1 α)} 1/ {α1 α)β1 β)} 1/ u 3 u 6 {α1 β)} 1/ {β1 α)} 1/) 69) {α1 β)} 1/ {β1 α)} 1/ {α1 α)β1 β)} 1/ u 3 u 6) 1/ u 3 Substituting 67) 69) into 66), we find that 1 1 ) 1/ u 3 ) 650) 1 u 3 u 6 ) 1/ u 1 u 3 u 6 ) 1/ u 3) ) 1/ Using Gröbner bases, A Strzebonski denested 650) obtained a polynomial of degree for u The value of u that we seek is a root of the factor u 3u 6 15u 3u 1 of this th degree polynomial If x u, then x 3x 3 15x 3x 1x x1/x) 3x 1/x)13 ) 0 Since x 1/x > 1, we find that Hence, x 1 x u 1 u 19 3, so that 1 u ) Lastly, we apply Lemma 33 with P u 3 Q G /G) 6 to deduce, from 651), that Q 1 Q u 3 u 3 ) u 1 u) 3 3u 1 u) ) 1 9 3) 1/ 179 3) Solving for 1/Q, we find that 65) 1 Q 1 9 3) 1/ 179 3) License or copyright restrictions may apply to redistribution; see

36 160 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG Thus, by 651) 65), G 11 Q 1/1 u 1/ 19 3) 1/ 179 3) ) 1/1 3 It remains to show that 19 3) 1/ 179 3) ) 3/ 7 ) 7, which is easily accomplished via Mathematica Second Proof of Theorem 5 We need two modular equations, one of degree 5 the other of degree 9 The first is found in Ramanujan s second notebook If β has degree 5 over α, then [1, p 1, Entry 13x)] 653) {α1 β)} 1/ {β1 α)} 1/ /3 {αβ1 α)1 β)} 1/ The second is found in Ramanujan s first notebook, but curiously not in his second R Russell [6] established this modular equation in 190, but his formulation is imprecise; in particular, it has a sign ambiguity We give Ramanujan s formulation Let P 1 αβ 1 α)1 β), { } Q 6 αβ 1 α)1 β) αβ1 α)1 β), 1/ R 3 αβ1 α)1 β) Then, if β has degree 9 over α, P P 17PR 1/3 9R /3 )R 1/6 9P Q 13PR 1/3 15R /3 ) 65) Let q exp π 9/5), so that we may apply 61) 6) with r 5 p9if u G 15 G 9/5 ) 1, then, by 653), {α1 β)} 1/ {β1 α)} 1/ {α1 β)} 1/ {β1 α)} 1/) 655) Thus, by 655), with α replaced by 1 α), {α1 α)β1 β)} 1/ u u 6 P 1 u u 6, Q 1u 6u 6 16u 1, R u 1 License or copyright restrictions may apply to redistribution; see

37 RAMANUJAN S CLASS INVARIANTS 161 Substitute these values into 65), square both sides, simplify, factor, with the help of Mathematica We then find that u 1)u u 1)u u 1)u 0u 6 3u 0u 1) u 1 9u 10 11u 16u 6 11u 9u 1) 0 In numerically checking the roots of each of these polynomials, we find that x : u is a root of x 0x 3 3x 0x 1x x1/x) 0x 1/x) 5 ) 0 Thus, x 1/x 10 15, so u 1/u 15 Hence, 1 u ) Lastly, we apply Lemma 3 with P u Q G 9/5 /G 15 ) 3 Then Q 1 ) ) 1 ) Q P P u u u u Hence, 1 Q ) From 656) 657), G 15 Q 1/6 u 1/ ) 1/ To complete the proof, we must show that ) 9 5 5) / Both equalities are easily verified Second Proof of Theorem 7 In addition to the modular equation of degree 3 given by 6), we need Ramanujan s modular equation of degree 71 If β has License or copyright restrictions may apply to redistribution; see

38 16 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG degree 71 over α, then [1, p, Entry 3ii)] 65) 1αβ) 1/ {1 α)1 β)} 1/ 1 1αβ) 1/ {1 α)1 β)} 1/)) 1/ αβ) 1/ {1 α)1 β)} 1/ {αβ1 α)1 β)} 1/ /3 {αβ1 α)1 β)} 1/ 1 αβ) 1/ {1 α)1 β)} 1/) Let r 3p 71 in equalities 61) principle 6) Thus, β has degree 71 over 1 α) Replacing α by 1 α in 65) employing 67) 69), but now with u G 13 G 71/3 ) 1, we deduce that 1 u 3 u 6) 1/ 1 1 ) 1/ u 3 ) 659) u 3 u 6) 1/ 1/ u 3) 1 u 3 ) u 1 u 3 u 6) 1/ 1/ u 3) u 3 u 6) ) 1/ 1/ u 3 1 u) 1 u 3 u Using resultants, Strzebonski M Trott denested 659) found a polynomial that factors into several polynomials of degrees, 1, Numerically eliminating all factors except one, we find that u satisfies u 0u 6 16u 0u 10 Letting u : x, solving for x 1/x, we find that x 1/x 0 3 It then follows that u 1/u 3Hence, 660) 1 u Lastly, we invoke Lemma 33 with P G 13 G 71/3 ) 3 u 3 QG 71/3 /G 13 ) 6 So, by Lemma ), Q 1 Q u 3 u 3 ) ) 1/ 1 3) Solving for 1/Q, we find that 1 Q 191 3) 1/ 1 661) 3) 5, , 00 3 License or copyright restrictions may apply to redistribution; see

39 RAMANUJAN S CLASS INVARIANTS 163 Hence, by 660) 661), G 13 Q 1/1 u 1/ ) 1/ 1 3) , , 00 ) 1/1 3 It thus remains to show that ) 1/ 1 3) 5, , ) 3/ ) 71, which can be verified via Mathematica 7 Watson s Empirical Process In [30], Watson employed an empirical process to evaluate 1 of Ramanujan s class invariants Motivated by Watson s idea, we succeeded in formulating theorems which give rigorous evaluations of G pq G p/q when p q are distinct primes satisfying pq 1mod)h pq) Let K Q m) msquarefree) be an imaginary quadratic field, let O K be its ring of integers By class field theory [13, p 191, Theorem 131], there exists an everywhere unramified extension K 1) K such that the Galois group GalK 1) K) C K The field K 1) is known as the Hilbert class field of K A Hilbert class field of K is usually defined as the maximal unramified abelian extension of K Let a [τ 1,τ ]beano K -ideal Define g 3 ja) j[τ 1,τ ]) 17 [τ 1,τ ]) g 3[τ 1,τ ]) 7g3 [τ 1,τ ]), where 1 g [τ 1,τ ]) 60 mτ 1 nτ ) g 3 [τ 1,τ ]) 10 m,n m,n) 0,0) m,n m,n) 0,0) 1/ 1 mτ 1 nτ ) 6 It is clear from the definitions of g [τ 1,τ ]) g 3 [τ 1,τ ]) that j[τ 1,τ ]) j[1,τ]) : jτ), where τ τ /τ 1 We will also let γ τ) 3 jτ) with the cube root being real-valued when ja) isreal License or copyright restrictions may apply to redistribution; see

40 16 BRUCE C BERNDT, HENG HUAT CHAN, AND LIANG CHENG ZHANG It is well known that K 1) KjO K )) [11, p 0, Theorem 111] If D K is the discriminant of K 3 D K,thenK 1) Kγ τ K )) [11, p 9, Theorem 1], where m, DK 0 mod ), τ K 3 m, D K 1 mod ) Lemma 71 Let a b be two O K -ideals Define σ a jb)) by 71) σ a jb)) jāb), where aā is a principal ideal Then σ a is a well-defined element of GalK 1) K), a σ a induces an isomorphism C K GalK 1) K) Proof See [11, p 0, Corollary 1137] Lemma 7 Let K Q pq), wherep q are two distinct primes satisfying pq 1mod), let {, if 3 pq, γ 1, if 3 pq Then G γ pq is a real unit generating the field K1) Proof From [5, p 90], we find that G 1 pq is a real unit of K1) Since [11, p 57, Theorem 117] jo K )j pq) 16G pq 7) )3, G pq we conclude that 73) K 1) KG 1 pq ) Next, suppose that 3 pq Then 3 D K γ τ K ) generates K 1) From the equality [11, p 57, Theorem 117] γ pq) 16G pq 7) G pq 73), we find that G pq K 1) Hence, G pq K 1), by 73) Remarks In [5, p 90], B J Birch quoted M Deuring s results [1, p 3] indicated that G pq is a unit when pq 1 mod ) A more elaborate proof of this statement can be found in [9, Corollary 5] In fact, from the treatment given in [9], one can show that G p/q is also a unit This fact will be needed in our main theorem From class field theory, we know that if H is a subgroup of C K, then there exists an abelian everywhere unramified extension L K such that GalK 1) L) H In particular, when H CK : the subgroup of squares in C K, the corresponding field M K is known as the genus field of K One can show that M is the maximal unramified extension of K which is abelian over Q [11, p 1] License or copyright restrictions may apply to redistribution; see