Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Distinguishing Hecke eigenforms Alex Ghitza The University of Melbourne
54th AustMS Conference Tuesday 28 September 2010
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Plan
1
Quick refresher on modular forms
2
A precise upper bound
3
Asymptotic results
4
How sharp is this? An experiment
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
For N ≥ 1, the congruence subgroup Γ0 (N) is defined as a b Γ0 (N) = ∈ SL2 (Z) : c ≡ 0 (mod N) . c d A modular form of weight k on the group Γ0 (N) is a holomorphic function f : H −→ C satisfying a growth condition at i∞ and az + b f = (cz + d)k f (z) cz + d a b for all z ∈ H, ∈ Γ0 (N). c d
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Modular forms have Fourier expansions of the form f (q) =
∞ X
an (f ) q n ,
q = e2πiz .
n=0
We say that f is cuspidal if a0 (f ) = 0. The spaces of modular forms have a family of commuting linear operators Tn . A simultaneous eigenvector for all the Tn ’s is called a Hecke eigenform. Every cuspidal eigenform f has a1 (f ) 6= 0, so it is customary to normalise f so that a1 (f ) = 1. The coefficients of a normalised eigenform f are algebraic integers in a number field Kf .
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
The question
How many initial Fourier coefficients can two distinct eigenforms f and g have in common? Theorem (Sturm 1987, Ram Murty 1997) Let f and g be distinct modular forms of the same weight k on the group Γ0 (N). Then there exists n ≤
k [SL2 (Z) : Γ0 (N)] such that an (f ) 6= an (g). 12
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
We are interested in the case of distinct weights. Theorem (G) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N). Then there exists n ≤ 4 (log(N) + 1)2 such that an (f ) 6= an (g). Note that the bound is independent of the weights.
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Murty’s idea Lemma (Ram Murty 1997) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N), and let p be the smallest prime not dividing N. Then there exists n ≤ p2 such that an (f ) 6= an (g). Proof. Suppose, on the contrary, that an (f ) = an (g) for all n ≤ p2 . The Fourier coefficients of f and g satisfy the recurrence relations ap2 (f ) = ap2 (f ) − pk1 −1 ,
ap2 (g) = ap2 (g) − pk2 −1 .
By assumption we have ap (f ) = ap (g) and ap2 (f ) = ap2 (g), so k1 = k2 , contradiction.
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Murty’s idea Lemma (Ram Murty 1997) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N), and let p be the smallest prime not dividing N. Then there exists n ≤ p2 such that an (f ) 6= an (g). Proof. Suppose, on the contrary, that an (f ) = an (g) for all n ≤ p2 . The Fourier coefficients of f and g satisfy the recurrence relations ap2 (f ) = ap2 (f ) − pk1 −1 ,
ap2 (g) = ap2 (g) − pk2 −1 .
By assumption we have ap (f ) = ap (g) and ap2 (f ) = ap2 (g), so k1 = k2 , contradiction.
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Proof of the Theorem Lemma (Ram Murty 1997) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N), and let p be the smallest prime not dividing N. Then there exists n ≤ p2 such that an (f ) 6= an (g).
Theorem (G) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N). Then there exists n ≤ 4 (log(N) + 1)2 such that an (f ) 6= an (g).
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Proof of the Theorem It suffices to prove that there exists p ≤ 2 log(N) + 2 not dividing N. If not, Y N≥ p. p≤2 log(N)+2
In terms of Chebyshev’s function θ(x) =
P
p≤x
log(N) ≥ θ(2 log(N) + 2). We then get a contradiction from the Lemma For all x ≥ 0, we have θ(2x + 2) > x.
log(p):
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Proof of the Theorem It suffices to prove that there exists p ≤ 2 log(N) + 2 not dividing N. If not, Y N≥ p. p≤2 log(N)+2
In terms of Chebyshev’s function θ(x) =
P
p≤x
log(N) ≥ θ(2 log(N) + 2). We then get a contradiction from the Lemma For all x ≥ 0, we have θ(2x + 2) > x.
log(p):
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Lemma For all x ≥ 0, we have θ(2x + 2) > x. The main ingredient is the following estimate: Theorem (Dusart’s PhD thesis, 1998) For all x > 1, |θ(x) − x| < 3.965 logx2 (x) . Important input in the proof of Dusart’s theorem: positions of the first 1.5 × 109 zeros of the Riemann zeta function, computed by Brent, van de Lune, te Riele, and Winter in 1986.
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
Theorem (G) Let f and g be cuspidal eigenforms of weights k1 6= k2 on the group Γ0 (N). Then there exists n ≤ log2 (N) + O log1.525 (N) (RH) n ≤ log2 (N) + O log1.5 (N)(log log N) (Cramér) n ≤ log2 (N) + O log(N)(log log N)2 such that an (f ) 6= an (g).
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
We again use Ram Murty’s lemma. The worst-case is given by the primorials Nk = p1 · · · pk , where the smallest prime not dividing Nk is pk +1 . Now use (un)conditional results on the gaps between primes:
pk +1
0.525 pk + O pk = pk + O pk0.5 log(pk ) pk + O log2 (pk )
(Baker-Harman-Pintz 2001) assuming RH (Cramér 1936) Cramér’s conjecture.
Quick refresher on modular forms A precise upper bound Asymptotic results How sharp is this? An experiment
What is the Truth? In level 1, our first theorem gives n ≤ 4. We took advantage of two level 1 artifacts (Victor Miller basis, Maeda’s conjecture) to run a numerical experiment in Sage comparing Fourier coefficients up to reasonably high weights. Assuming Maeda (the characteristic polynomial of T2 is irreducible), we checked that the coefficient a2 is sufficient to distinguish between cuspidal eigenforms of level 1 and weights up to 10,000. (We then checked the irreducibility of these polynomials up to weight 4,096, so this much is now unconditional.)
