SUPER-REPLICABLE FUNCTIONS N (j 1,N ) AND PERIODICALLY VANISHING PROPERTY

J. Korean Math. Soc. 44 007, No., pp. 343 371 SUPER-REPLICABLE FUNCTIONS N j 1,N AND PERIODICALLY VANISHING PROPERTY Chang Heon Kim and Ja Kyung Koo Reprinted from the Journal of the Korean Mathematical Society Vol. 44, No., March 007 c 007 The Korean Mathematical Society

J. Korean Math. Soc. 44 007, No., pp. 343 371 SUPER-REPLICABLE FUNCTIONS N j 1,N AND PERIODICALLY VANISHING PROPERTY Chang Heon Kim and Ja Kyung Koo Abstract. We find the super-replication formulae which would be a generalization of replication formulae. And we apply the formulae to derive periodically vanishing property in the Fourier coefficients of the Hauptmodul N j 1,1 as a super-replicable function. 1. Introduction Let H be the complex upper half plane and let Γ 1 N be a congruence subgroup of SL Z whose elements are congruent to 1 mod N N = 1,,.... Since the group Γ 1 N acts on H by linear fractional transformations, we get the modular curve X 1 N = Γ 1 N\H, as a projective closure of the smooth affine curve Γ 1 N\H, with genus g 1,N. Here, H denotes the union of H and P 1 Q. Ishida and Ishii showed in [11] that for N 7, the function field KX 1 N is generated over C by the modular functions X z, N ɛ N N and X 3 z, N N, r 1N 1 πi N 1 K where X r z, N = e r,s z 4N s=0 K and ɛ 1,sz N is 1 or according as N is odd or even. Here, K r,s z is a Klein form of level N for integers r and s not both congruent to 0 mod N. On the other hand, since the genus g 1,N = 0 only for the eleven cases 1 N 10 and N = 1 [5], [15], the function field KX 1 N in this case is a rational function field Cj 1,N for some modular function j 1,N Table 3, Appendix. The element 1 1 0 1 of Γ 1N takes z to z + 1, and in particular a modular function f in KX 1 N is periodic. Thus it can be written as a Laurent series in q = e πiz z H, which is called a q-series or q-expansion of f. We call f normalized if its q-series starts with q 1 + 0 + a 1 q + a q +. By a Hauptmodul t we mean the normalized generator of a genus zero function field KX 1 N and we write t = q 1 + 0 + k 1 H kq k for its q-series. For a Fuchsian group Γ, let Γ denote the inhomogeneous group of Γ = Γ/ ± I. Let Γ 0 N be the Hecke subgroup given by { c d SL Z c 0 Received August 1, 005. 000 Mathematics Subject Classification. 11F03, 11F. Key words and phrases. modular function, Hauptmodul, Thompson series, replicable, super-replicable. 343 c 007 The Korean Mathematical Society

344 CHANG HEON KIM AND JA KYUNG KOO mod N}. Also, let t = N j 1,N be the Hauptmodul of Γ 1 N and X n t be a unique polynomial in t of degree n such that X n t 1 n q n belongs to the maximal ideal of the local ring C[[q]]. Polynomials with this property are known as the Faber polynomials [5], Chapter 4. Write X n t = 1 n q n + m 1H m,n q m. When Γ 1 N = Γ 0 N, N j 1,N becomes a replicable function, that is, it satisfies the following replication formulae H a,b = H c,d whenever ab = cd and a, b = c, d [1], [3], []. Given a replicable function f the n-plicate of f is defined iteratively by f n nz = az + b f a + nx n f d ad=n 0 b<d where the primed sum means that the term with a = n is omitted [3]. We call f completely replicable if f is a replicable function with rational integer coefficients and has only a finite number of distinct replicates, which are themselves replicable functions. According to [1] there are, excluding the trivial cases q 1 + aq, 36 completely replicable functions of which 171 are monstrous functions, i.e., modular functions whose q-series coincide with the Thompson series T g q = n Z Trg V nq n for some element g of the monster simple group M whose order is approximately 8 10 53. Here we observe that V = n ZV n is the infinite dimensional graded representation of M constructed by Frenkel et al. [8], [9]. Furthermore, in [3] Cummins and Norton showed that if f is replicable, it can be determined only by the 1 coefficients of its first 3 ones. If Γ 1 N Γ 0 N, unlike those replicable functions mentioned above, we show in 3 that the Fourier coefficients of X n t with t = N j 1,N N 7, 9 satisfy a twisted formula 10 by a character ψ see Corollary 11. Here we note that when we work with the Thompson series, it is reduced to replication formulae in by viewing ψ as the trivial character. Thus in this sense it gives a more general class of modular functions, which we propose to call N j 1,N a super-replicable function. There would be certain similarity between some of replicable functions and super-replicable ones as follows. We derived in [19] the following self-recursion formulas for the Fourier coefficients of N j 1,N without the aid of its -plicate when N =, 6, 8, 10, 1: for k 1, H 4k 1 = H k 1 + 1 j k 1 + α H 4k H k 1 H j H 4k j 1 j k H j H 4k j

H 4k = β H 4k H 4k+1 = H k + SUPER-REPLICABLE FUNCTIONS 345 1 j<k H 4k+ = β H 4k 1 j<k 1 H j H k j 1 H j H 4k j + α H 4k + H k H j H k j, 1 j<k 1 j<k H j H 4k j where α = N j 1,N 1+N/ N and β = N j 1,N 1 N/. Furthermore, we verified in [0] that the above recursion can be also applied to 14 monstrous functions of even levels including N j 1, and N j 1,6 which are Thompson series of type B, 6C, 6E, 6F, 10B, 10E, 14B, 18C, 18D, B, 30C, 30G, 4C, 46AB these are all replicable functions and one monster-like function of type 18e for the definition of monster-like function, we refer to [6]. Therefore the Hauptmoduln mentioned above which have self-recursion formulas can be determined just by the first four coefficients H 1, H, H 3 and H 4 without the aid of -plicate. What is more interesting would be the fact that there seems to be a connection between super-replicable functions and infinite dimensional Lie superalgebras. That is, considering the arguments from Borcherds [], Kang [1] and Koike [] we have believed that the super-replication formulae in 10 might suggest the existence of certain infinite dimensional Lie superalgebra whose denominator identity implies such formulae. Meanwhile, Kang et al. [14] recently showed that there is indeed such a Lie algebra as follows. Let Γ be a free abelian group of finite rank and Γ be a countable usually infinite sub-semigroup in Γ such that every element in Γ can be written as a sum of elements of Γ in only finitely many ways. We consider a Γ Z -graded Lie superalgebra [13] with a product identity of the form exp 1 k νk α, ae kα,a = 1 ζβ, be β,b, α,a Γ Z k 1 β,b Γ Z where ν k α, a, ζβ, b Z and E λ,a are the basis elements of the semi-group algebra C[Γ Z ] with the multiplication given by E λ,a E µ,b = E λ+µ,a+b for λ, µ Γ and a, b Z. When rank Γ =, Γ= Z Z and Γ= Z >0 Z >0, they proved that one can get the product identity of the form exp 1 k νk m, np km q kn = 1 ζm, np m q n m,n=1 k 1 m,n=1 with p = E 1,0 and q = E 0,1, from which one derives a characterization of the super-replication formulae. Moreover, they further computed the supertraces of the Monster Lie superalgebras associated with super-replicable functions.

346 CHANG HEON KIM AND JA KYUNG KOO Lastly, as an application of super-replication formulae we consider the following periodically vanishing property. Many of monstrous functions, for example, Thompson series of type 4B, 4C, 4D, 6F, 8B, 8C, 8D, 8E, 8F, 9B, etc have periodically vanishing properties among the Fourier coefficients see the Table 1 in [4]. This result must be known to experts, but we could not find a reference. Hereby we describe it in Theorem 13. Meanwhile, as for the case of super-replicable functions, we see from the Appendix, Table 4 that only the Haupmodul N j 1,1 seems to have such property. To this end, we shall first derive in 4 an identity 4 which is analogous to the k -plication formula [7], [] satisfied by replicable functions. And, combining this with the superreplication formulae we are able to verify that the Fourier coefficients H m of N j 1,1 vanish whenever m 4 mod 6 Corollary 19. Through the article we adopt the following notations: S Γ1 N the set of Γ 1 N-inequivalent cusps q h = e πiz/h, z H f = f c d az+b cz+d fz = gz + O1 means that fz gz is bounded as z goes to i.. Hauptmodul of Γ 1 1 In this section we investigate the generalities of the modular function j 1,1 which is under primary consideration and construct the Hauptmodul N j 1,1. We also examine some number theoretic property of N j 1,1. As for more arithmetic properties, we refer to [10]. Lemma 1. Let a a c and c be fractions in lowest terms. Then a c is Γ 1N- equivalent to a c if and only if ± a c a+nc c mod N for some n Z. Proof. Straightforward. Using the above lemma we can check that the cusps 0, 1/, 1/3, 1/4, 1/5, 1/6, 1/8, 1/9,, 5/1 are Γ 1 1-inequivalent. But from [15] we know that the cardinality of S Γ1 1 is 10, whence S Γ1 1 = {0, 1/, 1/3, 1/4, 1/5, 1/6, 1/8, 1/9,, 5/1}. For later use we are in need of calculating the widths of the cusps of Γ 1 1. Lemma. Let a/c P 1 Q be a cusp where a, c = 1. Then the width of a/c in X 1 N is given by N/c, N if N 4. Proof. If N 4, the statement is obvious. Hence, we assume that N does not divide 4, i.e., N 1,, 4. First, choose b and d such that c d SL Z. Let

SUPER-REPLICABLE FUNCTIONS 347 h be the width of the cusp a/c. Then h is the smallest positive integer such that 1 1 h ±Γ c d 0 1 c d 1 N. Thus we have 1 cah c ±Γ h 1 + cah 1 N. 1 cah If c is an element of Γ h 1 + cah 1 N, by taking trace 1 cah mod N; hence N 4. Thus when N 1,, 4, c Γ h 1 + cah 1 N. This condition is equivalent to saying that h N c, N Z N ca, N Z = N c, N Z. We then have the following table of inequivalent cusps of Γ 1 1: Table 1. cusp 0 1/ 1/3 1/4 1/5 1/6 1/8 1/9 5/1 width 1 1 6 4 3 1 3 4 1 Recall the Jacobi theta functions θ, θ 3, and θ 4 defined by θ z = n Z θ 3 z = n Z q n+ 1 q n θ 4 z = 1 n q n n Z for z H. We have the following transformation formulas [8] pp.18-19. 1 3 4 5 6 θ z + 1 = e 1 4 πi θ z θ 3 z + 1 = θ 4 z θ 4 z + 1 = θ 3 z θ 1 = iz 1 θ4 z z θ 3 1 = iz 1 θ3 z z θ 4 1 = iz 1 θ z. z

348 CHANG HEON KIM AND JA KYUNG KOO Lemma 3. Let k be an odd positive integer and N be a multiple of 4. Then for F z M k Γ 0 N and m 1, F mz M k Γ 0 mn, χ m with χ m d = and d, m = 1. m d Proof. [31], Proposition 1.3. Put j 1,1 z = θ 3 z/θ 3 6z. Theorem 4. a θ 3 z M 1 Γ 0 4 and θ 3 6z M 1 Γ 0 1, χ 3. b KX 1 1 is equal to Cj 1,1 z. j 1,1 takes the following value at each cusp: j 1,1 = 1, j 1, = 3, j 1,1 1 = 0 a simple zero, j 1,1 1 3 = i, j 1,1 1 4 = 3i, j 1,1 1 5 = 3, j 1,1 1 6 = a simple pole, j 1,1 1 8 = 3i, j 1,1 1 9 = i, j 1,1 5 1 = 1. Proof. For the first part, we recall that [1], p.184 θ 3 z M 1 Γ 0 4. Then by Lemma 3 we immediately get that θ 3 6z M 1 Γ 0 1, χ 3. By the assertion a, it is clear that j 1,1 z KX 1 1. Thus for b, it is enough to show that j 1,1 z has only one simple zero and one simple pole on the curve X 1 1. As is well-known, θ 3 z never vanishes on H. Hence we are forced to investigate the zeroes and poles of j 1,1 at each cusp of Γ 1 1. Let S = 0 1 and T = 1 1 0 1. i s = : ii s = 0: θ 3 z j 1,1 = lim z i θ 3 6z = lim 1 + q + q 4 + q 0 1 + q 3 + q 1 + = 1. j 1, = θ 3 z lim z i = lim z i = 3. θ 3 6z S i z θ 3 z i z 6 θ 3 z 6 by 5 iii s = 1 : We Observe that ST S = 1 = 1.

SUPER-REPLICABLE FUNCTIONS 349 Considering the identities θ 3 z [S]1 = z 1 θ 3 = z 1 { iz 1 z θ3 } z by 5 = i z θ 3 θ 3 z [ST ] 1 = i z θ 3 [T ] 1 = i z θ 4 by 3 θ 3 z [ST S] 1 = i z θ 4 [S]1 = i z 1 { iz 1 θ z} by 6 we get that On the other hand = θ z, θ 3 z s= 1 = lim θ 3z [ST z i S] 1 = lim θ z z i = lim z i q 1 + q + q 6 + q 1 + since θ z = q 8 1 + q + q 3 + = 0 a triple zero. θ 3 z = 1 θ 3 z 1 1 3 3 0 0 1 1 1 1 = 1 θ 3 z 3 1 1 0 1 1 1 1 1 3 1 = 1 θ 3 z 1 1 3 1 3 1 = 1 3 θ z 1 1 0 3 1 = 3 1 θ z 1 = 3 1 e πi 3 q 6 1 +, 3 so that θ 3 6z has a simple zero at 1. Thus j 1,1 = θ3z θ 3 6z at 1, whence j 1,1 has a simple zero at 1. iv s = 1 3 : ST 3 S = 1 3. First we recall that θ 3 z = 3 θ 3 z θ 3 6z S θ 3 z 6 by 5. has a double zero Observe that θ z = 1 θ 3 z θ 4 z and θ 3z = 1 θ 3 z + θ 4 z. From these identities we can write θ 3 z = θ z + θ 3 z and θ 3 z 6 = θ 3 z +

350 CHANG HEON KIM AND JA KYUNG KOO θ 3 3z. Then we have that and θ 3 z = 3 θz + θ 3 z θ 3 6z ST 3 θ 3 z + θ 3 3 z T 3 = 3 = 3 πi e 4 6 θ z + θ 3 z e πi 4 θ 3 z + θ 3 3 z by 1, and 3, iθ z + θ 3 z iθ 3 z + θ 3 3 z θ 3 z = iθ z + θ 3 z 3 θ 3 6z ST 3 S iθ 3 z + θ 3 3 z S = iθ 4 z 3 + θ 3 z i 3θ 4 3 z + 3θ 3 3 z by 4 and 5 which goes to i+1 i+1 = i as z i, so that 1 j 1,1 = i. 3 v s = 1 4 : 1 4 0 1 = 1 4. In this case we use the following well-known fact from [1] p.148 : For γ Γ 0 4 and z H, where Θz = θ 3 z. Then c 1 1 Θγz = cz + d, d d θ 3 z Θ 1 4 0 1 = z θ 3 6z 4 1 Θ 3 0 0 1 1 4 0 1 z = Θ 4 1 z Θ 3 4 3 1 0 3 z 4z + 1 Θz = 4 3 i 1 4 z 3 + 3 Θ 1 0 3 z = 3i Θz Θ z 3 which tends to 3i when z goes to i. Therefore 1 j 1,1 = 3i. 4

SUPER-REPLICABLE FUNCTIONS 351 vi s = 1 5 : Because 4 5 5 1 0 1 sends to 1 5, j 1,1 1 5 θ 3 z = lim z i θ 3 6z 5 1 4 5 0 1 = lim θ 3z z i θ 3 6z by a 0 1 = j 1, = 3. vii s = 1 6 : Observe that ST 6 S = 1 6. Considering the identities θ 3 z = 3 θ3 z θ 3 6z S θ 3 z 6 by 5 θ 3 z = 3 θ3 z θ 3 6z ST 6 θ 3 z = 3 θ4 z θ 3 6z ST 6 S θ 4 z 6 = 3 S θ 3 z 6 = 3 θ4 z θ T 6 4 z 6 by and 3 iz θ z by 6, 6iz θ 6z we have that j 1,1 1 6 θ 3 z θ z = lim = lim z i θ 3 6z ST 6 z i S θ 6z q = lim 81 + q + q 6 + z i q8 61 + q6 + q 18 +. since θ z = q 8 1 + q + q 3 + Thus by Table 1, j 1,1 has a simple pole at 1 6. viii s = 1 8 : Because 5 1 36 7 4 1 sends to 1 8, j 1,1 1 8 θ 3 z = lim z i θ 3 6z 5 1 36 7 4 1 = lim θ 3z by a z i θ 3 6z 4 1 1 = j 1,1 = 3i. 4

35 CHANG HEON KIM AND JA KYUNG KOO ix s = 1 9 : Because 5 48 19 3 1 sends to 1 9, 1 θ 3 z j 1,1 = lim 9 z i θ 3 6z 5 48 19 3 1 = lim θ 3z by a z i θ 3 6z 3 1 1 = j 1,1 = i. 3 x s = 5 1 : 5 1 5 = 5 1. j 1,1 5 1 θ 3 z = lim z i θ 3 6z 5 1 5 = lim θ 3z by a z i θ 3 6z = 1. We will now construct the Hauptmodul N j 1,1 from the modular function j 1,1 mentioned in Theorem 4. j 1,1 z 1 = θ 3 6z θ 3 z θ 3 6z = 1 + q3 + q 1 + q 7 + q q 3 + q 4 + q 9 q 1 + = 1 q + q + q + q 3 q 6 q 7 q 8 q 9 + q 11 + q 1 +, which is in q 1 Z[[q]]. From the uniqueness of the normalized generator it follows that N j 1,1 = j 1,1 1. By Theorem 4-b we have the following table: Table. Cusp values of j 1,1 and N j 1,1 s 0 1/ 1/3 1/4 1/5 1/6 1/8 1/9 5/1 j 1,1 s 1 3 0 i 3i 3 3i i 1 N j 1,1 s 3 + 1 1 i 1 3i 1 3 0 1+ 3i 1 + i 1 Theorem 5. Let d be a square free positive integer and t = Nj 1,N be the normalized generator of KX 1 N. Let s be a cusp of Γ 1 N whose width is h s. If t q 1 Z[[q]] and s S Γ1 tz tshs is a polynomial in Z[t], N { } then tτ is an algebraic integer for τ Q d H. Proof. Let jz = 1 + 744 + 196884q +, the elliptic modular function. It q is well-known that jτ is an algebraic integer for τ Q d H [3], [30]. For algebraic proofs, see [4], [6], [9] and [3]. Now, we view j as a function

SUPER-REPLICABLE FUNCTIONS 353 on the modular curve X 1 N. Then j has a pole of order h s at the cusp s. On the other hand, tz ts has a simple zero at s. Thus j tz ts h s s S Γ1 N { } has a pole only at whose degree is µ N = [Γ1 : Γ 1 N], and so by the Riemann-Roch Theorem it is a monic polynomial in t of degree µ N which we denote by ft. Since s S Γ1 N { } tz tsh s is a polynomial in Z[t] and j, t have integer coefficients in the q-expansions, ft is a monic polynomial in Z[t] of degree µ N. This shows that tτ is integral over Z[jτ]. Therefore tτ is integral over Z for τ Q d H. Corollary 6. For τ Q d H, N j 1,1 τ is an algebraic integer. Proof. N j 1,1 has integral Fourier coefficients. And by Table 1 and, tz ts h s s S Γ1 1 { } = t t 1 t + 6 t + t + 4 t + t + 1 3 t t + 1 Z[t]. Now the assertion is immediate from Theorem 5. 3. Super-replication formulae Let n be the set of integral matrices c d, where a 1 + NZ, c NZ, and ad bc = n. Then n has the following right coset decomposition: See [1], [5], [30] n a 1 7 n i =, a n a,n=1 i=0 Γ 1 Nσ a a n 0 a where σ a SL Z such that σ a a 0 a mod N. Let fz be a modular function with respect to Γ 1 N. For brevity, let us call it fz is on Γ 1 N. For f KX 1 N we define an operator U n and T n by and n 1 f Un = n 1 f 1 i 0 n i=0 f Tn = n 1 a n a,n=1 n a 1 i=0 f σa a i 0 n a Lemma 7. For f KX 1 N and γ 0 Γ 0 N, f Tn γ0 = f γ0 Tn for any positive integer n. In particular, f Tn is again on Γ 1 N..

354 CHANG HEON KIM AND JA KYUNG KOO Proof. First we claim that Let c d n. Then γ 0 1 c d a n a,n=1 n γ 0 = γ 0 n γ0 = w y z x i=0 for γ 0 = x y z w Γ 0 N. c d x y z w d mod N. Hence γ n γ 0 n so that n γ 0 γ 0 n. By the same argument we can show the reverse inclusion. We note that n a 1 n a i γ 0 = Γ 1 Nσ a γ 0 and γ 0 n = = a n a,n=1 a n a,n=1 n a 1 i=0 n a 1 i=0 Here we note that σ a a i 0 n a f Tn γ0 and γ 0 σ a i a 0 n a Now the assertion follows. n 0 a γ 0 Γ 1 Nσ a a Γ 1 Nγ 0 σ a a n 0 a n 0 a i i because Γ 1 N Γ 0 N. γ 0 s are the matrices appearing in the definition of s are those appearing in the definition of f γ0 Tn. For a positive integer N with g 1,N = 0, we let t resp. t 0 be the Hauptmodul of Γ 1 N resp. Γ 0 N. And, we write X n t = 1 n q n + m 1 H m,nq m and X n t 0 = 1 n q n + m 1 h m,nq m. Lemma 8. For positive integers m and n, H m,n = H n,m and h m,n = h n,m. Proof. Let p = e πiy and q = e πiz with y, z H. Note that X n t can be viewed as the coefficient of p n -term in log p logty tz [7]. Thus H m,n becomes the coefficient of p n q m -term of log p logty tz = log1 p/q + logp 1 q 1 logty tz = 1 i p/qi F p, q, i 1 p where F p, q = log 1 q 1 + i 1 Hipi q i p 1 q. We then come up with F p, q 1 = F q, p, which implies that H m,n = H n,m. Similarly if we work with t 0 instead of t, the identity h m,n = h n,m follows.

SUPER-REPLICABLE FUNCTIONS 355 Theorem 9. For positive integers n and l such that n, N = l, n = 1, X l t Tn = X ln t σn + c, where c is a constant. In particular, t Tn = X n t σn + c. Proof. Since X l t has poles only at Γ 1 N, the poles of X l t Tn can occur 1 only at a i 0 n 1 σa Γ a 1 N, where a and i are the indices appearing in the definition of T n. On the other hand, we have 1 a i 0 1 n σa Γ a 1 N = n 1 n a i 1 σa Γ 0 a 1 N = n a i 1 σa Γ 1 N. Let γ be an element in Γ 1 N. Then 1 σa γ n n a i 0 a a i 0 a 0 a a 0 0 a 1 1 0 1 mod N n 0 1 = n + Nm Nk for some k, m Z. If there exists an integer x > 1 with x n + Nm, Nk, then x must divide det [ n a i 1 σa γ ] = n. In this case x N because n, N = 1. 0 a 1 Therefore 1 σa γ is of the form γ 0 for some γ 0 Γ 0 N. Now a i 0 n a we conclude that X l t Tn can have poles only at γ 0 for some γ 0 Γ 0 N. By Lemma 7, nx l t Tn γ0 = nx l t γ0 Tn = a n = a n a n n a 1 i=0 Here we note that X l t γ0 σa a i 0 n a n a X lt γ0 σa U na az = a n n a X lt γ0 σa U na az + X l t γ0 σ n nz. a n a n n a X lt γ0 σa U na az = O1. In fact, if γ 0 σ a / ±Γ 1 N, it is clear that X l t γ0 σa U na has a holomorphic q-expansion. Otherwise X l t γ0σ au n = X l t U na a = l 1 q l + terms of positive degree U na = O1

356 CHANG HEON KIM AND JA KYUNG KOO because l, n = 1. Now we have 8 nx l t Tn γ0 = X l t γ0σ n nz + O1. This implies that X l t Tn has a pole at γ 0 if and only if γ 0 σ n ±Γ 1 N, that is, γ 0 ±Γ 1 Nσ n 1. Hence X l t Tn has poles only at cusps Γ 1 Nσ n 1. In this case we derive from 8 nx l t Tn σn 1 = X l t σn 1 σ n nz + O1 = l 1 q ln + O1. Thus X l t Tn σn 1 = nl 1 q ln +O1 and X l t Tn σn 1 has poles only at cusps σ n Γ 1 Nσ 1 n = Γ 1 N. On the other hand, X ln t has poles only at Γ 1 N too and X ln t = ln 1 q ln + O1. Therefore X l t Tn σn 1 = X ln t + c for some constant c. Then we have X l t Tn = X ln t σn + c, as desired. Corollary 10. Let N be a positive integer such that the genus g 1,N is zero and [Γ 0 N : Γ 1 N]. For positive integers n, l and m such that n, N = l, n = 1, we have { } e 1 ψe H mn e,l h mn e,l + h mn e,l e m,n e>0 = ψnh m,ln h m,ln + h m,ln, where ψ : Z/NZ {±1} is a character defined by { 1, if e ±1 mod N ψe = 1, otherwise. Proof. It follows from Theorem 9 that 9 X l t Tn = e n e>0 e 1 X l t σe U ne ez = X ln t σn + constant. Note that for each positive integer r, { X r t, if e ±1 mod N X r t σe + X r t = X r t 0 + constant, otherwise. In the above when e is not congruent to ±1 mod N, X r t σe + X r t is on Γ 0 N and has poles only at Γ 0 N with r 1 q r as its pole part, which guarantees the above equality. We then have X r t σe = 1 {ψex rt X r t 0 + X r t 0 } + constant.

SUPER-REPLICABLE FUNCTIONS 357 Now 9 reads e n e>0 e 1 1 {ψex lt X l t 0 + X l t 0 } U ne ez = 1 {ψnx lnt X ln t 0 + X ln t 0 } + constant. Comparing the coefficients of q m -terms on both sides, we get the corollary. Corollary 11. Let N be a positive integer such that the genus g 1,N is zero and [Γ 0 N : Γ 1 N]. For positive integers a, b, c, d with ab = cd, a, b = c, d and b, N = d, N = 1, 10 H a,b = ψbdh c,d + 1 ψbd h c,d. Proof. In Corollary 10 we take n = b, l = 1 and m = a. Then it follows from the conditions and the replicability of h m,n that Now the assertion follows. ψbh a,b h a,b = ψdh c,d h c,d. Corollary 1. Let N be a positive integer with g 1,N = 0 and [Γ 0 N : Γ 1 N] =. If mn, N = 1 and mn ±1 mod N, then h m,n = H m,n. Proof. In Corollary 11 we take a = m, b = n and c = n, d = m. The condition that ψmn = 1 implies This proves the corollary. H m,n = H n,m + h n,m = H m,n + h m,n by Lemma 8. 4. Vanishing property in the Fourier coefficients of N j 1,1 As mentioned in the introduction, many of the Thompson series have periodically vanishing properties among the Fourier coefficients. Now we will give a more theoretical explanation for these phenomena. Let T g be the Thompson series of type g and Γ g be its corresponding genus zero group. To describe Γ g we are in need of some notations. Let N be a positive integer and Q be any Hall divisor of N, that is, Q be a positive divisor of N for which Q, N/Q = 1. We denote by W Q,N a matrix Qx y Nz Qw with det W Q,N = Q and x, y, z and w Z, and call it an Atkin-Lehner involution. Let S be a subset of Hall divisors of N and let Γ = N + S be the subgroup of P SL R generated by Γ 0 N = { c d SL Z c 0 mod N} and all Atkin-Lehner involutions W Q,N for Q S. For a positive divisor h of 4, let n be a multiple of h and set N = nh. When S is a subset of Hall divisors of n/h,

358 CHANG HEON KIM AND JA KYUNG KOO we denote by Γ 0 n h + S the group generated by h 0 1 Γ 0 n/h h 0 and h 0 1 W Q,n/h h 0 for all Q S. If there exists a homomorphism λ of Γ 0 n h + S into C such that 11 1 13 14 λγ 0 N = 1, λ 1 1/h 0 1 = e πi/h, { λ n 1 = e πi/h if n/h S, e πi/h if n/h / S, λ is trivial on all Atkin-Lehner involutions of Γ 0 N in Γ 0 n h + S, then we let n h + S be the kernel of λ which is a subgroup of Γ 0 n h + S of index h. Ferenbaugh [6] found out a necessary and sufficient condition for the homomorphism λ to exist and calculated the genera of groups of type n h + S. All the genus zero groups of type n h + S are listed in [6], Table 1.1 and 1.. Now we have the following theorem. Theorem 13. Write T g q = q 1 + m 1 c gmq m. i If Γ g = Γ 0 N and h is the largest integer such that h 4 and h N, then c g m = 0 unless m 1 mod h. ii If Γ g = N + S and there exists a prime p such that p N and p Q for all Q S, then c g m = 0 whenever m 0 mod p. iii If Γ g = n h + S, then c g m = 0 unless m 1 mod h. Proof. i Note that 1 h 1 0 1 belongs to the normalizer of Γ0 N. Thus T g 1 h has poles only at, which is a simple pole. This enables us to 1 0 1 write T g 1 h = c T 1 g for some constant c. By comparing the coefficients 0 1 of q m -terms on both sides, the assertion follows. ii From Corollary 3.1 [] it follows that T g Up = 0. Thus ii is clear. iii Considering the identity in [], p.7 we have T g z + 1/h = e πi/h T g z. Then T g q = q 1 + 0<l Z c glh 1q lh 1, which implies iii. Unlike the cases of Thompson series, when we handle the super-replicable function N j 1,1 we can not directly use the ingredients adopted in Theorem 13. Therefore we start with Lemma 14. For c d Γ0 1, j 1,1 the generalized quadratic residue symbol. c d Proof. Immediate from Lemma 3 and Theorem 4. = 3 d j1,1. Here denotes We fix N = 1 and let t denote the Hauptmodul N j 1,1 in what follows. Lemma 15. For c d Γ0 4, t U χ = 1 c 3 4 c d a + c t U χ, 4

SUPER-REPLICABLE FUNCTIONS 359 where χ = 1 is the Jacobi symbol and t U χ is the twist of t U by χ [1], p.17. Proof. From [1], p.18 we observe that t U χ = 1 t U z + 1 t U z + 3 15 4 4 4 = 1 1 t U + 1 4 i 4 i 0 4. i Z/4Z It then follows from [19], Corollary 8 that 1 t t t U = H. 6 If we compare the coefficients of q-term on both sides, we get H 4 t 1 6 H = 0. And, substituting H = 1 and H 4 = 0 we get t 1 6 = 0. Now 16 t U = 1 t = j 1,1 1. Then for c d Γ0 1 t U + 1 = 1 c d j 1,1 = 1 3 17 c d j 1,1 by Lemma 14 d 3 = t U + 1 3 = t U + 1 d a since ad 1 mod 1. Let c d Γ0 4. For i = 1, 3, we consider 4 0 4 i c d = 4a+ic 4b+id 4c 4d. Since 4a + ic, 4c divides det 4a+ic 4b+id 4c 4d, we must have 4a + ic, 4c = 4; hence a + ic/4, c = 1. Thus we can choose integers x i and y i such that γ i = a+ic/4 x i c y i SL Z. Write γ 1 i 4 0 4 i c d = 4 zi 0 4 for some integer zi. Then we have 4 0 4 i c d = a+ic/4 xi 4 zi c y i 0 4 for some integer z i. Comparing 1,-component on both sides we get Thus id a + ic/4z i 4b + id = a + ic/4z i + 4x i. mod 4. Then z i a + ic/4id mod 4, because n 1 mod 4 for every odd integer n aid + i c 4 d mod 4 i + c d mod 4, due to ad 1 mod 4 4 i + 6c 1 d mod 4, where we write c = 4c 1.

360 CHANG HEON KIM AND JA KYUNG KOO Therefore we derive that for c d Γ0 4, 4 t U χ c d 1 = t U + 1 i = = i Z/4Z i Z/4Z i Z/4Z 1 i 1 i 4 i 0 4 by 15 c d t U + 1 a+ic/4 x i 4 zi c y i 0 4 3 a + ic/4 t U + 1 Now if we set c = 4c 1 as before, then we have 4 t U χ c d 1 3 = i a + 6c 1 i i Z/4Z 3 1 = a + 6c 1 i i Z/4Z 4 z i by 17. 0 4 t U + 1 4 i+6c 1 d 0 4 t U + 1 4 i+6c 1d 0 4 since a + 6c 1 i a + 6c 1 mod 1 3 1 1 1 = a + 6c 1 i i + 6c 1 d i + 6c 1 d i Z/4Z t U + 1 4 i+6c 1 d 0 4 3 = 1 c1 4 t U a + 6c χ 1 because 1 1 1 1 i i+6c 1d = i +6c 1id = 1+c 1id = 1 c1id = 1 c1 and i + 6c 1 d runs over Z/4Z. This completes the lemma. Lemma 16. i For each k 1, we put gz = 1 k 1 1 Then g belongs to KX 1 4. ii For c d Γ0 4 and k 1, t Uk χ c d In particular, t lies in KX U k χ 14. X ktz X kt z + 1. = 1 c 3 4 a + c t. U k χ 4

SUPER-REPLICABLE FUNCTIONS 361 Proof. First we note that for n N, T n = U n. Here, by n N we mean that n divides some power of N. To show g KX 1 4, we observe that g = 1 k 1 X ktz X kt U z. Then using Lemma 7 we obtain that g KX 1 4. For c d 18 t U k + 1 = c d Γ0 1, t U + 1 U k 1 c d = t U + 1 by Lemma 7 U c d k 1 3 = t U + 1 U by 17 k 1 a 3 = t U k a + 1. Now we can proceed in the same manner as in the proof of Lemma 15. Lemma 17. For k 1, i t U k 6 1 = ii t U k 3 1 = { 1 q 1 + O1, if k = 1 O1, otherwise. O1, if k = 1 1 4 q 4 1 + O1, if k = 8 q 1 + O1, if k = 3 1 k 1 i q k 4 + O1, if k 4. k Proof. i First, t 6 1 is holomorphic at because t has poles only at the cusps Γ 1 1. Now for k 1, t Uk 6 1 = t U U k 1 6 1 = 1 t Uk 1 0 1 6 0 1 + 1 = 1 t Uk 1 1 1 0 + 1 = 1 z t Uk 1 + 1 by Lemma 7 { 1 = q 1 + O1, if k = 1 O1, otherwise. t Uk 1 1 1 t 7 4 1 7 U k 1 0 6 1 t Uk 1 7 4 1 1 0 1 7 1 1 0

36 CHANG HEON KIM AND JA KYUNG KOO ii We observe that t 3 1 O1. And for k 1, 19 t Uk 3 1 = t U U k 1 3 1 = 1 = 1 t Uk 1 0 3 1 + 1 t Uk 1 6 1 0 + 1 t Uk 1 1 1 0 3 1 t Uk 1 1 3 0 0 1 has a holomorphic Fourier expansion if k = 1. Thus we suppose k. We then derive that t Uk 1 1 3 = t U U k 1 = 1 = 1 = 1 t Uk 3 0 1 3 + 1 t Uk 3 1 1 0 1 + 1 t Uk 1 1 0 1 3 t Uk 11 1 1 1 6 1 1 1 t Uk 3 1 0 1 1 + 1 t Uk 6 1 1. If we substitute the above into 19, for k, 0 0 1 t Uk = 1 + 1 4 = 1 3 1 t Uk 1 t Uk t Uk 1 6 1 0 + 1 4 3 1 1 0 1 0 6 1 0 + 1 4 t Uk 6 1 1 0 0 1 0 1 t Uk 6 1 1 0 When k =, + 1 4 t Uk 3 1 4 0 1 1. t U4 3 1 = 1 z t U 6 1 + 1 4 t 1 6 0 1 z + 1 + 1 4 t 1 3 0 1 4z + 1 = 1 4 q 4 1 + O1 by i.

SUPER-REPLICABLE FUNCTIONS 363 If k = 3, t U8 3 1 = 1 z t U 4 6 1 + 1 4 t U 6 1 z + 1 + 1 4 t U 3 1 4z + 1 = 1 1 8 e πiz+ i + O1 = 8 q 1 + O1 by i and the case k = 1 in ii. For k 4, we will show by induction on k that 3 t Uk 3 1 = i 1k 1 k k 4 q + O1. First we note that by i and 0 If k = 4, t Uk 3 1 = 1 4 t Uk 3 1 4z + 1 + O1. t U4 3 1 = 1 t U4 4 3 1 4z + 1 + O1 = 1 πi e 4z+1 + O1 by the case k = 16 = 1 4 1 i 4 4 4 q + O1. Thus when k = 4, 3 holds. Meanwhile, if k = 5 then we get that t U5 3 1 = 1 t U5 4 3 1 4z + 1 + O1 = i 3 e πi4z+1 + O1 by the case k = 3 = 1 5 1 i 5 5 4 q + O1. Therefore in this case 3 is also valid. Now for k 6, t Uk 3 1 = 1 4 This proves the lemma. Theorem 18. For k 1, 4 3 1 4z + 1 + O1 i e πi4z+1 k by induction hypothesis for k t Uk = 1 4 1k 1 = 1 k 1 t Uk χ z = 1k 1 1 i k k 4 q + O1. k 4 + O1 X ktz X kt z + 1.

364 CHANG HEON KIM AND JA KYUNG KOO This identity twisted by a character χ is analogous to the k -plication formula [7], [] satisfied by the Hauptmodul of Γ 0 N. Proof. Put gz = 1 k 1 1 X ktz X kt z + 1 as before. We see by Lemma 16 that both t and g sit in KX U k χ 14. For t U = g, we k χ will show that t g has no poles in U k χ H. Recall that t Uk χ = 1 4 = 1 4 i Z/4Z χi i Z/4Z Since t has poles only at Γ 1 1, t U k k 1 j=0 χ t Uk 4 i 0 4 1 4 0 4 i 1 1 j 0 k Γ1 1 = 16 1 k 4 i 0 4 Let c d Γ1 1. Then k+ 4j i 0 4 c d = = k+ 4j i 0 4 k+ a 4jc ic 4c χit 1 j 0 k 4 i 0 4. can also have poles only at k j 0 1 Γ 1 1. Γ 1 1 = k+ a 4jc ic. 4c Observe that k+ a 4jc ic, 4c k 16 = k+4. Write k+ a 4jc ic, 4c = l for some integer l 0. Then s = k+ a 4jc ic/ l 4c/ l is in lowest terms. Since 1 c, s is of the form s = 3m for some integers m and n. We assume 3m, n = 1. Here we consider two cases. i m: Choose integers x and y such that 3m n x y SL Z and consider t Uk χ 3m n x y = 1 t 4 U k 4 0 1 4 3m n x y t U k 4 0 3 4 3m n x y = 1 t 4 U 4n+3m t k U 4n+9m. k 1m 1m Since 3m, n = 1 and m, we can write 4n+3m 1m = γ0 U 1 and 4n+9m 1m = γ 0 U, where both γ 0 and γ 0 are in Γ 0 1 and U 1, U are upper triangular matrices. Then by 18, t U k χ 3m n x y O1. Hence, if m then n t Uk is holomorphic at the cusp s = χ 3m. ii m: n

SUPER-REPLICABLE FUNCTIONS 365 If 3 m, then s is of the form s = 3m n x y with 3m n x y Γ 0 4. Thus by Lemma 16, t Uk χ 3m n x 3m 3 y = 1 4 t U k n + 3m/4 O1. χ As for the other cases, if we use Lemma 1, it is easy to see that s is equivalent to 1 1 or 5 1 under Γ 14. Thus we conclude that t can have poles only at 1 U k χ 1, 5 1 under Γ 14- equivalence. Next, let us investigate the poles of g. Recall that X kt has poles only at Γ 1 1. Therefore g can have poles only at 0 i 1 Γ 1 1 for i = 0, 1. And, for c d Γ1 1, 0 i 1 c d = i a ic 0 c d = c = a ic/ in lowest terms. c Hence by Lemma 1, g can have poles only at 1 4, 5 4, 7 4, 11 4, 1 1, 5 equivalence. At 1 4, 5 4, 7 4 For example, at 5 4, and 11 4 1 under Γ 14-, it is easy to check that g is holomorphic. g 5 1 4 5 = 1k 1 1 X kt 5 1 4 5 X kt 1 = 1 k 1 1 X kt 5 1 4 5 X kt 34 48 10 7 = 1 k 1 1 0 5 1 4 5 X kt 5 1 4 5 X kt 17 1 4 17 1 0 O1 since 5 4, 17 4 / Γ 11. Now it remains to show that t U g has no poles at the cusps equivalent k χ to 1 1, 5 1 under Γ 14. At 1 1, t Uk χ 1 1 = 1 5 t 4 U k 4 0 1 4 1 1 t U k 4 0 3 4 1 1 = 1 t 4 U k 1 3 0 1 16 1 0 1 t U 5 1 k 6 1 8 1 0 = 1 t 4 U k 1 3 0 1 16 1 0 1 t U 11 1 k 1 1 = 1 4 t U k 3 1 16z + 1 t U k 6 1 6 1 8 0 1 4z + 1.

366 CHANG HEON KIM AND JA KYUNG KOO By 5 and Lemma 17, we have the following: If k = 1, t U χ 1 1 = 1 1 1 4 e πi4z+ + O1 If k =, t U = 1 4 q + O1. 4 χ 1 1 = 1 1 πi e 16z+1 + O1 4 If k = 3, If k 4, = 1 8 q 4 + O1. t U3 χ 1 1 = 1 i 4 8 e πi16z+1 + O1 t Uk = 1 4 χ 1 k 1 = 1 16 q 8 + O1. 1 1 i k e πi k 4 16z+1 + O1 = 1 k 1 1 k q + O1. k+1 Observe that the identities for k = 1, and 3 are the same as the last one when k 4. Hence we conclude that for all k 1, t Uk χ 1 1 = 1 k 1k 1 q + O1. k+1 On the other hand, g 1 1 = 1k 1 1 X kt 1 1 X kt 0 1 1 1 = 1 k 1 1 X kt X kt 7 4 1 7 1 0 = 1 k 1 1 k q + O1. k+1 Thus t Uk χ g 1 1 O1. At 5 1, we see that 5 1 5 = 19 48 5 5 1 1 sends to 1. Then t Uk = t U k = 1 48 4 χ 5 1 5 χ 19 48 5 1 1 3 19 + 48 4 t U k χ 1 1 by Lemma 16

And g 5 1 5 = 1k 1 1 SUPER-REPLICABLE FUNCTIONS 367 = 1 t U k χ 1 1 = 1 k 1 k q + O1. k+1 = 1 k 1 1 X kt 5 1 5 X kt 0 1 5 X kt 5 1 5 X kt 11 1 1 1 1 0 = 1 k 1 X kt 0 1 + O1 = 1 k 1 k q + O1. k+1 This implies that t Uk χ g 5 1 5 O1, from which the theorem follows. 1 5 Now, we are ready to show periodically vanishing property of N j 1,1. Corollary 19. As before we let t be the Hauptmodul of Γ 1 1 and write X n t = 1 n q n + m 1 H m,nq m. Then we have i H m, k = 1 k 1 1 H k m,1 for odd m. ii H m = 0 whenever m 4 mod 6, and m = 5. Proof. First if we compare the coefficients of q m -terms on both sides of the identity in Theorem 18, we get i. We see from the Appendix, Table 4 that H 5 = 0. On the other hand, by the super-replication formula Corollary 11 it follows that for m relatively prime to 1, { H H m, k = H k,m = m,1, if m ±1 mod 1 k H m,1, if m ±5 mod 1 k because h k m,1 = 0 in this case [], Corollary 3.1. Then H k m,1 = 0 when k is odd and m 5 mod 6, or k is even and m 1 mod 6. It is easy to see that { k m k, m 1, k odd, m 5 mod 6} { k m k, m 1, k even, m 1 mod 6} = {l Z l 4 mod 6}. This proves ii. Appendix. Fourier coefficients of the Hauptmodul N j 1,N We shall make use of the following modular forms to construct j 1,N. For z H, ηz the Dedekind eta function G z Eisenstein series of weight

368 CHANG HEON KIM AND JA KYUNG KOO G p z = G z pg pz for each prime p E z = G z/ζ normalized Eisenstein series of weight E p z = E z pe pz for each prime p P N,a z = P a1 z+a Lz N N-th division value of P, where a = a 1, a, L z = Zz + Z and P L τ is the Weierstrass P-function relative to a lattice L Now we get the following tables due to [16]-[18]: Table 3. Hauptmoduln N j 1,N N j 1,N N j 1,N 1 jz jz 744 θ z 8 /θ 4 z 8 56/j 1, + 4 3 E 4 z/e 4 3z 40/j 1,3 1 + 9 4 θ z 4 /θ 3 z 4 16/j 1,4 8 5 4ηz 5 /η5z+e 5 z η5z 5 /ηz 8/j 1,5 + 44 5 G z G 3z G 6 /j 1,6 1 1 z G3 z P 7 7,1,0 7z P 7,,0 7z P 7,1,0 7z P 7,4,0 7z 1/j 1,7 1 3 8 θ 3 z/θ 3 4z /j 1,8 1 1 P 9,1,0 9z P 9,,0 9z 9 P 9,1,0 9z P 9,4,0 9z 1/j 1,9 1 P 10 10,1,0 10z P 10,,0 10z P 10,1,0 10z P 10,4,0 10z 1/j 1,10 1 1 θ 3 z/θ 3 6z /j 1,1 1 When N = 1,, 3, 4, 6, N j 1,N becomes a Thompson series T g with Γ g = Γ 0 N. Hence, if N = 4, N j 1,4 has periodically vanishing property by Theorem 13-i. Otherwise, the Fourier coefficients of N j 1,N do not vanish see [4], Table 1. Therefore, we consider only the following cases N for which Γ 1 N Γ 0 N. Table 4. Fourier coefficients H m of N j 1,N for 1 m 60 N j 1,5 N j 1,7 N j 1,8 N j 1,9 N j 1,10 N j 1,1 H 1 10 4 3 1 H 5 3 1 1 H 3-15 0 1 1 1 1 H 4-4 -5 - - 0 H 5 15-7 -4 - - H 6 70 - -4-3 - -1 H 7 30 8 0 - - -1 H 8-15 16 6 1-1 -1 H 9-175 1 9 4 1-1 H 10 95-7 8 6 3 0

SUPER-REPLICABLE FUNCTIONS 369 N j 1,5 N j 1,7 N j 1,8 N j 1,9 N j 1,10 N j 1,1 H 11 40-9 -1 5 4 1 H 1 180-35 -1 1 4 H 13-615 -10-0 -5 1 H 14-86 37-16 -11 - H 15 410 70 1-1 -6 1 H 16 1760 53-7 -8 0 H 17 705-1 38 3-7 - H 18-415 -106 30 15-3 -3 H 19-3100 -16 1 4-4 H 0 1530-38 -40 19 10-4 H 1 670 119-64 5 14 - H 460 6-5 -15 H 3-8090 164 - -3 6 3 H 4-10174 -70 68-36 -6 5 H 5 4840-36 107 - -16 7 H 6 19570-378 88 8-6 H 7 7500-106 - 40-0 4 H 8-4360 353-11 58-8 0 H 9-3004 65-180 50 8-4 H 30 14130 469-144 1 6-8 H 31 55970-189 3-41 34-10 H 3 1155-885 18-84 31-9 H 33-67380 -1015 9-93 1-6 H 34-8196 -90 8-54 -14 0 H 35 37895 910 4-41 6 H 36 148410 1664-86 103-54 1 H 37 55305 1179-45 148-47 14 H 38-174500 -483-356 14-0 14 H 39-09577 -05-4 3 3 8 H 40 9605-49 440-96 6 H 41 37160-69 686-00 84-10 H 4 137160 1 544-19 7-18 H 43-47665 3998-5 -18 31 - H 44-508800 809-668 46-3 -0 H 45 30670-110 -1044 31-90 -1 H 46 885070-5119 -816 330 - H 47 33605-5754 5 75-107 15 H 48-1001340 -1598 996 67-44 6 H 49-118113 499 1563-16 45 33 H 50 531545 8968 110-439 133 9

370 CHANG HEON KIM AND JA KYUNG KOO N j 1,5 N j 1,7 N j 1,8 N j 1,9 N j 1,10 N j 1,1 H 5670 651 6-477 174 19 H 5 734130-506 -1464-75 154 0 H 53-53515 -1185-76 107 61-0 H 54-639348 -1579-1768 501-68 -37 H 55 1178880-3455 -8 708-19 -45 H 56 4456650 1081 18 590-54 -4 H 57 1606500 1978 384 146-0 -6 H 58-490150 1336 55-447 -90 0 H 59-5703676 -578-9 -911 100 7 H 60 5370-3765 -3056-987 7 5 References [1] D. Alexander, C. Cummins, J. Mckay, and C. Simons, Completely replicable functions, In: Groups, Combinatorics and Geometry, Cambridge Univ. Press 199, 87 98. [] R. E. Borcherds, Monstrous moonshine and monstrous Lie superalgebras, Invent. Math. 109 199, no., 405 444. [3] C. J. Cummins and S. P. Norton, Rational Hauptmoduls are replicable, Canadian J. Math. 47 1995, no. 6, 101 118. [4] M. Deuring, Die Typen der Multiplikatorenringe elliptischer Funktionenkörper, Abh. Math. Sem. Univ. Hamburg 14 1941, 197 7. [5] P. L. Duren, Univalent Functions, Grundlehren der Mathematischen Wissenschaften, 59, Springer-Verlag, 1983. [6] C. R. Ferenbaugh, The genus-zero problem for n h-type groups, Duke Math. J. 7 1993, no. 1, 31 63. [7], Replication formulae for n h-type Hauptmoduls, J. Algebra 179 1996, no. 3, 808 837. [8] I. B. Frenkel, J. Lepowsky, and A. Meurman, A natural representation of the Fischer- Griess monster with the modular function J as character, Proc. Nat. Acad. Sci. U. S. A. 81 1984, no. 10, Phys. Sci., 356 360. [9], Vertex operator algebras and the monster, Pure and Applied Mathematics, 134, Academic Press Inc., Boston, MA, 1988. [10] K. J. Hong and J. K. Koo, Generation of class fields by the modular function j 1,1, Acta Arith. 93 000, no. 3, 57 91. [11] N. Ishida and N. Ishii, The equation for the modular curve X 1 N derived from the equation for the modular curve XN, Tokyo J. Math. 1999, no. 1, 167 175. [1] S. J. Kang, Graded Lie superalgebras and the superdimension formula, J. Algebra 04 1998, no., 597 655. [13] S. J. Kang and J. H. Kwon, Graded Lie superalgebras, supertrace formula, and orbit Lie superalgrbra, Prod. London Math. Soc. 3 800, no. 3, 675 74. [14] S. J. Kang, C. H. Kim, J. K. Koo, and Y. T. Oh, Graded Lie superalgebras and superreplicable functions, J. Algebra 85 005, no., 531 573. [15] C. H. Kim and J. K. Koo, On the genus of some modular curve of level N, Bull. Austral. Math. Soc. 54 1996, no., 91 97. [16], Generators of function fields of the moldular curves X 1 5 and X 1 6, preprint. [17], Arithmetic of the modular function j 1,8, Ramanujan J. 4 000, no. 3, 317 338. [18], Generation of Hauptmoduln of Γ 1 7, Γ 1 9 and Γ 1 10, preprint.

SUPER-REPLICABLE FUNCTIONS 371 [19], Self-recursion formulas satisfied by Fourier coefficients of some modular functions, J. Pure Appl. Algebra 160 001, no. 1, 53 65. [0], Self-recursion formulas of certain monstrous functions, J. Pure Appl. Algebra 170, no. 1, 7 40. [1] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Graduate Texts in Mathematics, 97, Springer-Verlag, 1984. [] M. Koike, On replication formula and Hecke operators, Nagoya University preprint. [3] S. Lang, Elliptic Functions, Graduate Texts in Mathematics, 11, Springer-Verlag, 1987. [4] J. Mckay and H. Strauss, The q-series of monstrous moonshine and the decomposition of the head characters, Comm. Algebra 18 1990, no. 1, 53 78. [5] T. Miyake, Modular Forms, Springer-Verlag, 1989. [6] A. Néron, Modèles minimaux des variétés abéliennes sur les corps locaux et globaux, Publ. Math. I. H. E. S. 1 1964, 5 18. [7] S. P. Norton, More on moonshine, In: Computational Group Theory, Academic Press 1984, 185 193. [8] R. Rankin, Modular Forms and Functions, Cambridge University Press, 1977. [9] J. P. Serre and J. Tate, Good reduction of abelian varieties, Ann. Math. 88 1968, 49 517. [30] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton Univ. Press, 1971. [31], On modular forms of half integral weight, Ann. Math. 97 1973, 440 481. [3] J. H. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Graduate Texts in Mathematics, 151, Springer-Verlag, 1994. Chang Heon Kim Department of mathematics Seoul Women s university Seoul 139-774, Korea E-mail address: chkim@swu.ac.kr Ja Kyung Koo Korea Advanced Institute of Science and Technology Department of Mathematics Taejon 305-701, Korea E-mail address: jkkoo@math.kaist.ac.kr