A Note on the Transcendence of Zeros of a Certain Family of Weakly Holomorphic Forms

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1 A Note on the Transcendence of Zeros of a Certain Family of Weakly Holomorphic Forms Jennings-Shaffer C. & Swisher H. (014). A Note on the Transcendence of Zeros of a Certain Family of Weakly Holomorphic Forms. International Journal of Number Theory 10() 09. doi:10.114/s /S World Scientific Publishing Co. Accepted Manuscript

2 A NOTE ON THE TRANSCENDENCE OF ZEROS OF A CERTAIN FAMILY OF WEAKLY HOLOMORPHIC FORMS CHRIS JENNINGS-SHAFFER AND HOLLY SWISHER Abstract. Recently Duke and Jenkins have studied a certain family of modular forms f km that form a natural basis of the space of weakly holomorphic modular forms of weight k on SL (Z). In particular they prove that for all but at most k + 1 values of m the zeros of these functions all lie on the unit circle. 1 Using a method of Kohnen we observe that other than i ρ these zeros are transcendental. In addition we observe that in the remaining cases there are at most finitely many algebraic zeros and examine the values of these algebraic zeros in a special case. 1. Introduction Recall that a weakly holomorphic modular form of even integer weight k on the modular group Γ = PSL (Z) is a holomorphic function f : H C where H is the upper half of the complex plane such that for all τ H and ( a b c d ) Γ f ( ) aτ + b = (cτ + d) k f(τ). cτ + d In addition f is meromorphic at i.e. f has a Fourier expansion of the form f(τ) = n n 0 a n q n where n 0 = ord (f) Z and throughout q = e πiτ. We write M k! to denote the infinite dimensional complex vector space of weakly holomorphic modular forms of weight k and M k (resp. S k ) the corresponding finite dimensional space of holomorphic modular (cusp) forms of weight k. In [] Duke and Jenkins study a certain family of weakly holomorphic modular forms f km M k! of even weight k = 1l + k where l is an integer and k }. These functions have the property that for integers m l the Fourier expansion of f km has the form f km (τ) = q m + a km (n)q n. Furthermore the set f km m l} is a basis for M! k. n l+1 An interesting feature of these functions is that the coefficients satisfy the following duality [] a km (n) = a km (n). Zeros of certain classes of f km have been well-studied and shown to lie on the unit circle. For example in [1] Asai Kaneko and Ninomiya prove this when k = 0. In [] Duke and Jenkins provide a general proof that when m l l the zeros of f km all lie on the unit circle. 010 Mathematics Subject Classification. Mathematics Subject Classification 010: 11F11 11J81. Key words and phrases. Number theory; modular forms; zeros of modular forms; transcendental numbers. 1

3 1.1. Transcendence of zeros of modular forms. Recall the standard fundamental domain of the action of Γ on H is given by } 1 F = z H : Re(z) 0 z 1 z H : 0 < Re(z) < 1 } z > 1. Due to the valence formula which states that for a nonzero modular form f of weight k on Γ (1) k 1 = ord (f) + 1 ord i(f) + 1 ord ρ(f) + ord τ (f) τ F τ iρ the algebraic points i ρ = e πi F are frequently zeros of modular forms. For k 4 even let E k M k (Γ) denote the classical Eisenstein series of weight k E k (τ) = 1 + k σ k 1 (n)q n B k n 1 where B k is the k th Bernoulli number and σ k 1 (n) = d n dk 1. In 00 Kohnen [9] showed in an elegant argument that other than i and ρ the zeros of E k in F are transcendental. In addition he shows that the weakly holomorphic modular forms defined by J n = (j 744) T (n) where T (n) is the usual nth Hecke operator have this property as well. In fact the J n are polynomials in j with integer coefficients and J n (τ) = f 0n (τ). Kohnen s argument uses the beautiful result of F. Rankin and Swinnterton-Dyer [11] which states that the zeros of E k all lie on the following arc of the unit circle () A = e iθ : π θ π In 006 Gun [8] uses Kohnen s technique together with work of Getz [6] [7] to obtain similar results for another class of modular forms. Here we apply this method to show that in general other than i ρ the zeros of f km in F are transcendental when m l l. In particular we prove the following. Theorem 1.1. Let k be an even integer and write k = 1l+k such that k }. If m l l and z 0 is a zero of f km in the fundamental domain F then either z 0 i ρ} or z 0 is transcendental. The case when m = 0 is given by Gun in [8]. By the results of Kohnen [9] stated above we note the following corollary which follows directly from the proof of Theorem 1.1. Corollary 1.. Consider a weakly holomorphic modular form f(τ) of the form i= }. N M f(τ) = (τ) a E i (τ) bi P k (j(τ)) c k where b i c k are nonnegative integers a is any integer and P k are any polynomials in j(τ) with integer coefficients such that the zeros of P k all lie in [0 178]. Then other than i ρ all zeros of f(τ) in the fundamental domain F of Γ are transcendental. In Section we give a proper definition for the functions f km and review some results from class field theory. In Section we give the proof of Theorem 1.1. In addition we give the following result about algebraic zeros and examine the values of these zeros in a special case in Section 4. Theorem 1.. Let k be an even integer and write k = 1l + k such that k }. If l m < l l and z 0 is a zero of f km in the fundamental domain F then either z 0 Z D or z 0 is transcendental. Here D = l + m and Z D is a finite set dependent only on D. In Theorem 1. one should note for fixed D there are infinitely many corresponding f km. k=1

4 . Preliminaries Recall the classical Delta function (τ) S 1 (Γ) is defined by (τ) = E 4(τ) E 6 (τ) 178 and has the following infinite product expansion (τ) = q (1 q n ) 4. In addition the j-function j(τ) M! 0(Γ) defined by has a simple pole at. n=1 j(τ) = E 4(τ) (τ) For an even integer k let k = 1l + k where l is an integer and k }. For each integer m l there exists a unique weakly holomorphic modular form of weight k call it f km such that the q-expansion of f km is of the form f km (τ) = q m + O(q l+1 ). In particular f km = l E k F kd (j) where F kd (x) is a monic polynomial in j(τ) of degree D = l + m with integer coefficients. The polynomials F kd (x) are generalized Faber polynomials [4] [5]. In [] Duke and Jenkins prove that for m l l all zeros of f km in F lie on the arc A = e iθ : π θ π } and in particular that f km has D simple zeros on the interior of A. The proof of Theorem 1.1 utilizes the following lemma of Schneider which can be found in [10] (Corollary.4). Lemma.1. (Schneider 197) If z H and j(z) is algebraic then either z is transcendental or z is imaginary quadratic i.e. Q(z) is a degree extension of Q with z R..1. Some results from class field theory. We now recall some basic facts from class field theory and complex multiplication discussed in []. Consider an integer D < 0 so that K = Q( D) is an imaginary quadratic field. An order O of K is a subring of K containing 1 that is a free Z-module of rank. In addition a proper fractional ideal of O is a non-zero fractional ideal A of O such that () O = α K : αa A}. The set of all proper fractional ideals of K forms a multiplicative group with many nice properties. Consider a polynomial P (x) = ax + bx + c of negative discriminant D = b 4ac with integer coefficients such that a > 0 and gcd(a b c) = 1. If z H is a root of P (x) then as seen in [] (Lemma 7.5) O = [1 az] is an order of K and L = [1 z] is a proper fractional ideal of O. To see the structure of O we notice that since z H is a root of the polynomial ax + bx + c by the quadratic formula z = b+ D a. Thus [ [1 az] = 1 b + ] [ D 1 ] D b is even = [ ] b is odd D Since D = b 4ac we have that b is even if and only if D 0 (mod 4). Similarly b is odd if and only if D 1 (mod 4). We have thus shown the following.

5 Lemma.. Let a b c Z such that a > 0 gcd(a b c) = 1 and D = b 4ac < 0. If z H is a root of the polynomial ax + bx + c then the lattice [1 z] is a proper fractional ideal of the order O = [1 az] of K = Q( D). Moreover [ ] D 0 (mod 4) O = [ 1 D 1 1+ D ] D 1 (mod 4). In light of Lemma. we see that the order O doesn t depend on z but instead on the discriminant D of the reduced integer polynomial that has z as a root. Recall if L a lattice of C we define j(l) = j(z) where z H and L = [1 z] (the choice of z H is well defined up to Γ-equivalence). From Lemma. we see we can map a point z H to the proper fractional ideal L = [1 z] of O where j([1 z]) = j(z). The following lemma follows from Theorem 11.1 and Proposition 1. in [] and is the last result we need before the proof of Theorem 1.1. Lemma.. If A is a proper fractional ideal of an order O of an imaginary quadratic field K then j(a) is an algebraic integer over Q. If B is any other proper fractional ideal of O then K(j(A)) = K(j(B)) and j(a) and j(b) are conjugate over K. Furthermore the degree of j(a) is the class number of O.. The Proof of Theorem 1.1 Proof. Let k be an even integer and write k = 1l + k such that l is an integer and k }. Recall f km (τ) = (τ) l E k (τ)f kd (j(τ)) where F kd (x) is a degree D = m + l polynomial with integer coefficients. By Kohnen [9] the only possible zeros of E k (τ) are i and ρ. Also we see from the valence formula (1) that (τ) is never zero on H. Thus the only zeros of f km (τ) in F other than i ρ are the zeros of F kd (j(τ)). Suppose z 0 F such that F kd (j(z 0 )) = 0. Since F kd (x) is a polynomial with integer coefficients j(z 0 ) is algebraic. Thus by Lemma.1 z 0 is either transcendental or imaginary quadratic. If z 0 is imaginary quadratic then z 0 is a root of a polynomial P (x) = ax + bx + c where gcd(a b c) = 1 a > 0 and the discriminant D 0 = b 4ac < 0. Let K = Q( D 0 ). Consider the order O = [1 az 0 ] of K. By Lemma. the lattice [1 z 0 ] is a proper fractional ideal of O and the order O has the form [ 1 ] D 0 D 0 0 (mod 4) (4) O = [ ] 1 1+ D 0 D 0 1 (mod 4). Thus by Lemma. if A is any other proper fractional ideal of O j(z 0 ) = j([1 z 0 ]) and j(a) are conjugate. Keeping this in mind consider the point z 1 C defined by i D0 z 1 = D 0 0 (mod 4) D 0 D 0 1 (mod 4). Then z 1 F and from (4) we have [1 z 1 ] = O. Thus by definition [1 z 1 ] is a proper fractional ideal of O and so j(z 0 ) and j(z 1 ) are conjugate. 4

6 We may now take an automorphism σ of K(j(O)) such that σ(j(z 0 )) = j(z 1 ). Since σ acts as the identity on Q and F kd is a polynomial with integer coefficients we have that 0 = σ(0) = σ(f kd (j(z 0 ))) = F kd (σ(j(z 0 ))) = F kd (j(z 1 )). Thus z 1 is also a zero of F kd and hence a zero of f km. Since z 1 F by Duke and Jenkins [] we have that z 1 must lie on the arc of the unit circle given by e iθ : π θ π }. Suppose D 0 0 (mod 4) so that D 0 = 4n for some positive integer n. Then z 1 = i n but since z 1 must lie on the unit circle we must have n = 1. Thus D 0 = 4. Since z 0 H we have by the quadratic formula that b + i z 0 =. a But z 0 F and so Im(z 0 ). Thus a = 1 and so z 0 = b + i. But again by Duke and Jenkins [] we have that z 0 must lie on the unit circle so b = 0 and z 0 = i. If D 0 1 (mod 4) then D 0 = 4n + 1 for some positive integer n. Hence z 1 = 1 + i 4n 1 and thus z 1 = n. Again since z 1 must lie on the unit circle we must have n = 1. Thus D 0 =. Since z 0 H we have by the quadratic formula that z 0 = b + i. a And again since z 0 F this forces a = 1 so that z 0 = b + i. But again by Duke and Jenkins [] we have that z 0 must lie on the unit circle so b = 1 and z 0 = ρ. Thus a zero z 0 F of f mk is either transcendental or is one of i or ρ. 4. Remaining Cases Here it is worth noting that this result does not hold for all f km. As remarked by Duke and Jenkins [] while the f km are considered for all m l the zeros need not lie on the arc unless m l l and indeed one can compute counter examples Proof of theorem 1.. Proof. In the proof of Theorem 1.1 without knowing something about the location of z 1 we can only say that z 0 is either transcendental or imaginary quadratic and j(z 0 ) is conjugate to j(z 1 ). However we know j(z 1 ) = j(o) j(z 1 ) to be of degree given by the class number of the order O and there are only finitely many orders with a fixed class number. Thus for fixed D and varying k we may occasionally have an algebraic zero of F kd but there are only finitely many such zeros the rest are transcendental. 5

7 4.. An example. With k = 0 and D = 1 one verifies that F 1l1 (x) = x (744 4l). Thus for 4l > 744 or 984 < 4l the zero of F 1l1 (x) is outside [0 178] and so f 1l1 l has a zero in the fundamental domain off the arc as observed in []. Here we are considering orders with class number 1. From [] we have these are exactly from the following values: D 0 z 1 j(z 1 ) -1 i 1 - i 0 - ρ i i i Thus we see the zero of F 1l1 (j(τ)) to be algebraic for l = and transcendental otherwise. References [1] T. Asai M. Kaneko and H. Ninomiya. Zeros of certain modular functions and an application. Comment. Math. Univ. St. Paul. 46(1): [] D. Cox. Primes of the form x + ny. A Wiley-Interscience Publication. John Wiley & Sons Inc. New York Fermat class field theory and complex multiplication. [] W. Duke and P. Jenkins. On the zeros and coefficients of certain weakly holomorphic modular forms. Pure Appl. Math. Q. 4(4 Special Issue: In honor of Jean-Pierre Serre. Part 1): [4] G. Faber. Über polynomische Entwickelungen. Math. Ann. 57(): [5] G. Faber. Über polynomische Entwicklungen II. Math. Ann. 64(1): [6] J. Getz. A generalization of a theorem of Rankin and Swinnerton-Dyer on zeros of modular forms. Proc. Amer. Math. Soc. 1(8): [7] J. Getz. Corrigendum to A generalization of a theorem of Rankin and Swinnerton-Dyer on zeros of modular forms [mr0597]. Proc. Amer. Math. Soc. 18(): [8] S. Gun. Transcendental zeros of certain modular forms. Int. J. Number Theory (4): [9] W. Kohnen. Transcendence of zeros of Eisenstein series and other modular functions. Comment. Math. Univ. St. Pauli 5(1): [10] S. Lang. Introduction to transcendental numbers. Addison-Wesley Publishing Co. Reading Mass.-London-Don Mills Ont [11] F. K. C. Rankin and H. P. F. Swinnerton-Dyer. On the zeros of Eisenstein series. Bull. London Math. Soc. : Department of Mathematics University of Florida 58 Little Hall PO Box Gainesville Florida 611 USA address: cjenningsshaffer@ufl.edu Department of Mathematics Oregon State University 68 Kidder Hall Corvallis Oregon 971 USA address: swisherh@math.oregonstate.edu 6