STURM-LIKE BOUND FOR SQUARE-FREE FOURIER COEFFICIENTS PRAMATH ANAMBY AND SOUMYA DAS A BSTRACT. In this short article, we show the existence of an analogue of the classical Sturm’s bound in the context of the square–free Fourier coefficients for cusp forms of square-free levels. This number is a cut-off to determine a cusp form from its initial few square–free Fourier coefficients. We also mention some questions in this regard.

1. I NTRODUCTION The theory of modular forms by now occupies a central place in number theory, and its wide ranging applications in various branches of mathematics is well known. One pleasant, and computationally important feature of these objects is that if f is such a form in Mk (Γ) (Γ ⊆ SL2 (Z) is a congruence subgroup and k ≥ 0) with a Fourier expansion, say ∞

(1.1)

f (τ) =

(τ ∈ H = {z ∈ C | ℑ(z) > 0})

∑ a( f , n)e2πinτ ,

n=0

then there exists a number A > 0 depending on the space such that if a( f , n) = 0 for all n ≤ A, then f = 0. The smallest such bound in general is known as Sturm’s bound in the literature. Let us denote it by µ(k, Γ) and recall that (1.2)

µ(k, Γ) :=

k [SL2 (Z) : Γ] . 12

In fact Sturm’s bound is known for various kinds of modular forms, e.g., half-integer weight forms, Siegel modular forms etc. In this paper we will discuss the following question. Suppose that Γ = Γ0 (N) and consider the spaces Sk (N, χ) where χ is a Dirichlet character modulo N with conductor mχ and k is an integer. — Question. Let f ∈ Sk (N, χ) have the Fourier expansion (1.1). Suppose N/mχ is square–free. Does there exist a number B > 0, depending only on k, N such that if a( f , n) = 0 for all n ≤ B and n square–free, then f = 0? 2010 Mathematics Subject Classification. Primary 11F30. Key words and phrases. Sturm bound, square free, Fourier coefficients. 1

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PRAMATH ANAMBY AND SOUMYA DAS

We stress that the question has a trivial solution of f is a newform, in which case one can take B = µ(k, Γ0 (N)) with n ≤ B to be prime. The above question was motivated by recent work of the authors, who proved the following result. Theorem 1 ([3]). Let χ be a Dirichlet character of conductor mχ and let N be a positive integer with mχ |N such that N/mχ is square-free. Suppose that f ∈ Sk (N, χ) and that a( f , n) = 0 for all but finitely many square-free integers n. Then we have f = 0. The above mentioned Question is now a natural, finite counterpart to the above theorem. Note that the condition on the ratio of the level and conductor is necessary, this can be seen by taking the example of a non-zero form g(τ) ∈ Sk (SL2 (Z)) and considering g(m2 τ) for some m > 1. We would like to mention that before Theorem 1, the possibility of the existence of such a constant B as above, didn’t cross our minds. In this paper, we show the existence of B, and prove a crude bound for it in terms of k, N. We believe, with some efforts these bounds can be improved, and indicate some avenues for improvements. Let us define µsf (k, N) to be the smallest integer such that whenever f ∈ Sk (N, χ) with N/mχ square–free and a( f , n) = 0 for all square-free n ≤ µsf (k, N), then f = 0. Theorem 2. Let N be square–free and k ≥ 2. Then µsf (k, N) exists and satisfies the bound 2 (7k2 N)

µsf (k, N) ≤ a0 · N · 2r(r−1)/2 e4r log where a0 is an absolute constant and r =

,

(k−1)N . 2

The plan of the paper is as follows. In section 2 we prove the main result of the paper, and also include some conditional results. The proof follows an argument of Balog and Ono from [1] and it uses the existence of primes p is suitable intervals such that a0 ( f , p) 6= 0, where we put k−1

a0 ( f , n) := a( f , n)n− 2 . Moreover we also need such primes which distinguish between two newforms, say f1 6= f2 ; i.e., a( f1 , p) 6= a( f2 , p) with the p’s distinct and as small as possible. In section 2.1, we mention how to improve upon the various bounds by using the prime number theorem in short intervals for automorphic representations or by assuming a form of Maeda’s conjecture in the case of SL2 (Z). In the last section, several related questions are discussed. Acknowledgements. The second named author thanks the organisers of the conference “Lfunctions and Automorphic forms” held in Heidelberg in February 2016 for their kind invitation, for providing the opportunity to write this article and for their warm hospitality. He also thanks IISc., Bangalore and UGC centre for advanced studied for financial support. The authors thank the referee for comments on the paper.

STURM-LIKE BOUND

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2. P ROOF OF THE MAIN RESULT The following lemma follows immediately from the Prime Number Theorem, but we need to keep track of the dependence on the analytic conductors. Lemma 2.1. There exists C = C(N, k) > N such that for all X ≥ C, there exists a prime p ∈ (X, 2X] for which (i) given any newform f of any level M | N contained in Sk (N, χ), one has a( f , p) 6= 0; (ii) given any two distinct newforms f , g as in (i), one has a( f , p) 6= a(g, p). Proof. For X ≥ 1, define ψ f ×g (X) = ∑n≤x Λ f ×g (n), where we put −L0 ( f ⊗ g, s)/L( f ⊗ g, s) =

∞

∑ Λ f ×g(n)n−s.

n=1

By the prime number theorem for f ⊗ g, one has (see [5, Theorem 5.13]) that 1/2

|ψ f ×g (X) − r f ,g X| ≤ c2 q f ,g X exp{−c1 (log(X)1/2 )}

(2.1)

where c1 , c2 > 0 are absolute constants, q f ,g is the analytic conductor of the automorphic representation attached to f ⊗ g (see [5, § 5.1]), and r f ,g is the order of the possible pole (or zero) at s = 1. For future reference, let us note here that q f ,g ≤ 34 q2f · q2g .

(2.2)

k+1 and that qh = M( k−1 2 + 3)( 2 + 3), if h is a newform of level M.

Moreover since N is square–free, the Euler-factors of L( f ⊗ g, s) behave nicely, and an easy estimate (see [5, (5.49)]) shows that for an absolute constant c3 , |ψ f ×g (X) −

(2.3)

∑ a0f (p)a0g(p) log(p)| ≤ c3 X 1/2 log2(Xq f ,g).

p≤X

Let us call the quantities on the right-hand side of (2.1) and (2.3) to be R1 (X) and R2 (X) respectively and put R(X) := |R1 (X)| + |R2 (X)| and F := f − κg, with f , g as in the theorem and κ ∈ {0, 1}. Taking into account that r f ,g ≤ 1, and equality holds if and only if g = f . It is immediate from the above that

∑

|a0F (p)|2 log(p) = |a0f (p)|2 log(p) + κ|a0g (p)|2 log(p) − 2κℜ{a0f (p)a0g (p) log(p)}

X
(2.4)

≥ X − 4(R(X) + R(2X)).

A simple calculation shows that (2.4) is positive provided X > c4 exp{4 log2 (k2 N)}, where c4 is an absolute constant. So the lemma follows by taking C = c5 exp{4 log2 (7k2 N)} with c5 chosen so that C > N. 

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The rest of the section is devoted to a proof of Theorem 2. If f ∈ Sk (N, χ) is a newform then the result follows from the multiplicativity of the Fourier coefficients and the fact that for every prime p, the Hecke operators T (pν ) are polynomials in T (p) and clearly for this f , B = µ(k, Γ0 (N)) works. The following is adapted from Balog-Ono [1]. So let f ∈ Sk (N, χ) be non-zero. Consider the set { f1 , f2 , ...... fs } of all newforms of weight k and level dividing N contained in Sk (N, χ). Let their Fourier expansions be given by fi (z) = n ∑∞ n=1 bi (n)q . Then for all primes p, one has Tp f i = bi (p) f i . By "multiplicity-one", if i 6= j, we can find infinitely many primes p > N such that bi (p) 6= b j (p). Now by the theory of newforms, there exists αi,δ ∈ C such that f (z) has can be written uniquely in the form s

f (z) = ∑ ∑ αi,δ fi (δ z).

(2.5)

i=1 δ |N

Since f 6= 0, we may, after renumbering the indices, assume α1,δ 6= 0 for some δ |N. Let C be as in Lemma 2.1. Let p1 be any prime such that p1 ∈ (2C, 22C] and b1 (p1 ) 6= b2 (p1 ). Note that n (p1 , N) = 1. Then consider the form g1 (z) = ∑∞ n=1 a1 (n)q := Tp1 f (z) − b2 (p1 ) f (z) so that s

g1 (z) = ∑ (bi (p1 ) − b2 (p1 )) ∑ αi,δ fi (δ z). δ |N

i=1

The cusp forms f2 (δ z) for any δ | N, do not appear in the decomposition of g1 (z) but f1 (δ z) does for some δ |N. Also it is easy to see that a1 (n) = a( f , p1 n) + χ(p1 )p1k−1 a( f , n/p1 ) − b2 (p1 )a( f , n). Proceeding inductively in this way, and choosing the primes pi ∈ (2iC, 2i+1C] (2 ≤ i ≤ s − 1), we can remove all the non-zero newform components fi (δ z) for all i = 2, ..., s, to obtain a cusp form F(z) in Sk (N, χ). After dividing by a suitable non-zero complex number we get ∞

F(z) =

∑ A(n)qn := ∑ α1,δ f1(δ z).

n=1

δ |N

Now by repeating the above steps we get finitely many algebraic numbers β j and positive rational numbers γ j such that for every n (2.6)

A(n) =

∑ α1,δ b1(n/δ ) = ∑ β j a( f , γ j n).

δ |N

j

Let δ1 be the smallest divisor of N such that α1,δ1 6= 0 in (2.5). Define S = {p : p prime, p|N} ∪ {p : p prime, b1 (p) = 0}. Let us choose p ∈ (C, 2C] as in Lemma 2.1 (i). Then p ∈ / S, and A(δ1 p) = α1,δ1 b1 (p) 6= 0. Note that δ1 is square–free and (δ1 , γ j p) = 1 when γ j p ∈ N. From (2.6), we get a γ j ∈ N such

STURM-LIKE BOUND

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that a( f , δ1 γ j p) 6= 0. Now combining the estimates δ1 ≤ N,

γj ≤ 2

s(s−1) 2 −1Cs−1 ,

p ≤ 2C,

we obtain n ≤ B such that a( f , n) 6= 0. Moreover B satisfies µsf (k, N) ≤ B ≤ 2

s(s−1) 2 NCs .

Since N is square-free, s ≤ (k − 1)N/2 (see [6]). The theorem follows by substituting the expressions for s and C. This completes the proof of Theorem 2.  2.1. Remarks. In this section, we discuss the number µsf (N, k) and remark on a few ways of improving its bound, mostly based on some conjectures and numerical evidence. (i) An inequality. When Γ = SL2 (Z), by employing the so-called Miller’s basis, we can easily show that µsf (k, 1) ≥ µ(k, 1). (ii) A conjecture. Based on standard heuristics about Fourier coefficients of cusp forms and that the square–free integers have a positive natural density, we are led to believe that for all N ≥ 1, µ(k, N) ≤ µsf (k, N) ≤ a · µ(k, N) for some absolute constant a > 1. Numerical experiments with SAGE supports this. (iii) Application of a form of Maeda’s Conjecture. Let us recall a result due to P. Bengoechea [2] which states that if k ≡ 0 mod 4, and if for some n ≥ 1 the characteristic polynomial Tn,k (X) of the n-th Hecke operator Tn on SL2 (Z) is irreducible over Z and has full Galois group, then so are Tp,k (X) for all primes p. So, if we assume the aforementioned condition for some n, an inspection of the proof of Theorem 2 shows that we can choose p, p1 , p2 , . . . ,ps−1 to be consecutive primes in increasing order. We easily get that µsf (k, N)  exp logµ(k,1) µ(k,1) , where the implied constant is absolute. In particular, due to computations by Ghitza and Mc Andrew [4], this holds for all such k ≤ 12000. (iv) Primes in short intervals. If instead of the prime number theorem in long intervals that we used, one uses a short interval version for the PNT for the Rankin-Selberg convolution π ⊗ πe where π is the irreducible unitary representation obtained from a newform using [9], there is a possibility of improving the bound in Theorem 2, but we do not know the dependence of the implied constants on π in the above mentioned result. 2.2. Further questions. Throughout this section we assume N/mχ is square–free, k ≥ 2. (i) Eisenstein series. Even though we have stated the result only for cusp forms, it should be true for the space of Eisenstein series as well.

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PRAMATH ANAMBY AND SOUMYA DAS

(ii) Different method of proof . In order to prove something like (ii) in section 2.1, one should come up with a more natural method of proof, which does not involve choosing primes distinguishing between modular forms. Let us mention here that in [3], apart from Theorem 1, an asymptotic formula for the quantity ∑n≤X, n square–free |a0 ( f , n/t)|2 was shown, where t is a certain square–free integer which depends on N and the implied constants depend on k, N and f . We do not see immediately how one can deduce from there an answer to the above Question uniformly for all f ∈ Sk (N, χ). (iii) Finite version. Let Mk (N, OF ) denote the space of modular forms in Mk (N) whose Fourier expansion at ∞ lies in OF , the ring of integers of a number field F. Fix a prime ideal P ⊂ OF . In analogy with Sturm’s bound for finite primes (see e.g., [7]), is it true (with standard notations) that if f , g ∈ Mk (N, OF ) such that a( f , n) ≡ a(g, n) mod P for all square–free n ≤ µsf (k, N), then f ≡ g mod P? (iv) Half-integral weight forms. Given a result of A. Saha [8] (which holds e.g., in Kohnen’s + + plus space Sk+1/2 (4N) of level 4N with N odd, square–free) that f ∈ Sk+1/2 (4N) are determined by Fourier coefficients which are indexed by fundamental discriminants, one can ask for the existence of a finite ‘fundamental’ version of our result.

R EFERENCES [1] A. Balog, K. Ono: The Chebotarev density theorem in short intervals and some questions of Serre. J. Number Theory., 91, (2001), no.2, 356-371. [2] P. Bengoechea: On the irreducibility and Galois group of Hecke polynomials. arXiv:1703.02840. [3] P. Anamby, S. Das: Distinguishing Hermitian cusp forms of degree 2 by a certain subset of all Fourier coefficients. Preprint. [4] A. Ghitza, A. McAndrew: Experimental evidence for Maeda’s conjecture on modular forms. Tbil. Math. J. 5 (2012), no. 2, 55-69. [5] H. Iwaniec, E. Kowalski: Analytic Number Theory. Am. Math. Soc, Colloquium Publications, 53 (2004). [6] G. Martin: Dimensions of the spaces of cusp forms and newforms of Γ0 (N) and Γ1 (N). J. Number Theory., 112 (2005), 298-331. [7] M. R. Murty: Congruences between modular forms, in Analytic Number Theory, (ed. Y. Motohashi), London Mathematical Society Lecture Notes 247 (1997) 313-320, Cambridge University Press. [8] A. Saha: Siegel cusp forms of degree 2 are determined by their fundamental Fourier coefficients. Math.Ann., (2013), 363-380. [9] R. J. L. Oliver, J. Thorner: Effective log-free zero density estimates for automorphic L-functions and the Sato-Tate conjecture. arXiv:1505.03122.

STURM-LIKE BOUND

D EPARTMENT OF M ATHEMATICS , I NDIAN I NSTITUTE OF S CIENCE , BANGALORE – 560012, I NDIA . E-mail address: [email protected], [email protected] D EPARTMENT OF M ATHEMATICS , I NDIAN I NSTITUTE OF S CIENCE , BANGALORE – 560012, I NDIA . E-mail address: [email protected], [email protected]

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