Modular Degrees of Elliptic Curves

and

Discriminants of Hecke Algebras

William Stein^{∗}

            http://modular.fas.harvard.edu

ANTS VI, June 18, 2004

∗This is joint work with F. Calegari.

Goal

Let p be a prime. My goal is to explain and justify the following Calegari-Stein conjectures (note: 3 implies 2 implies 1):

Conjecture 1: If E/Q is an elliptic curve of conductor p, then the modular degree m_{E} of E is not divisible by p.

Conjecture 2: If T_{2}(p) is the Hecke algebra associated to S_{2}(p), then p does not divide the index of T_{2}(p) in its normalization.

Conjecture 3: If p ≥ k − 1, then there is an explicit formula for the p-part of the index of T_{k}(p) in its normalization.

Conj 1: If E of conductor p_{E}, then

p_{E} ∤ m_{E}.

A Motivation: Conjecture 1 looks like Vandiver’s conjecture, which asserts that p ∤ h_{p}^{−}. Flach proved the modular degree annihilates � (Sym ^{2}(E)), which is an analogue of a class group.

Conj 1: If E of conductor p_{E}, then

p_{E} ∤ m_{E}.

Watkins Data: For p_{E} < 10^{7} there are 52878 curves of prime conductor whose modular degree Watkins computed. No counterexamples to Conjecture 1 in the data. There are 23 curves such that m_{E} is divisible by a prime ℓ > p_{E}. For example the curve y^{2} + xy = x^{3} − x^{2} − 391648x − 94241311 of prime conductor p_{E} = 4 847 093 has modular degree 2 · 21 695 761. Smallest p_{E} with some ℓ > p_{E} is p_{E} = 1 194 923.

More Data

• The maximum known ratio \frac{m_{E}}{p_{E}} is ∼ 23.2, attained for p_{E} = 7 944 197.

• First curve with \frac{m_{E}}{p_{E}} > 1 has p_{E} = 13723 and m_{E} = 16176 = 2^{4} · 3 · 337.

• Smallest known \frac{m_{E}}{p_{E}} > 1 is 1.0004067 … for p_{E} = 1 757 963 where m_{E} = p_{E} + 715.

Modular Forms

Congruence Subgroup:

Γ_{0}(N) = \{(\begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(Z) such that N | c\} .

Cusp Forms: S_{k}(N) = {f : 𝔥 → C such that f(γ(z)) = (cz + d)^{−k}f(z) all γ ∈ Γ_{0}(N),

and f is holomorphic at the cusps  }

Fourier Expansion: f = _{n≥1}^{∑} a_{n}e^{2πizn} = _{n≥1}^{∑} a_{n}q^{n} ∈ C[[q]].

Computing Modular Forms

S_{k}(N) = 0 if k is odd, so we will not consider odd k further.

For k ≥ 2, a basis of S_{k}(N) can be computed to any given precision using modular symbols. Appears that no formal analysis of complexity has been done. Certainly polynomial time in N and required precision. Is polynomial factorization over Z the theoretical bottleneck?

Implemented in MAGMA

> S := CuspForms(37,2);
> Basis(S);
    q + q^3 - 2*q^4 - q^7 + O(q^8),
    q^2 + 2*q^3 - 2*q^4 + q^5 - 3*q^6 + O(q^8)
See also http://modular.fas.harvard.edu/mfd

Basis for S_{14} (11):

> S := CuspForms(11,14); SetPrecision(S,17);
> Basis(S);
q   - 74*q^13 - 38*q^14 + 441*q^15 + 140*q^16 + O(q^17),
q^2 - 2*q^13 + 78*q^14 + 24*q^15 - 338*q^16 + O(q^17),
q^3 + 18*q^13 - 72*q^14 + 89*q^15 + 492*q^16 + O(q^17),
q^4 + 12*q^13 + 31*q^14 - 18*q^15 - 193*q^16 + O(q^17),
q^5 - 10*q^13 + 46*q^14 - 63*q^15 - 52*q^16 + O(q^17),
q^6 + 11*q^13 - 18*q^14 - 74*q^15 - 4*q^16 + O(q^17),
q^7 - 7*q^13 - 16*q^14 + 42*q^15 - 84*q^16 + O(q^17),
q^8 - q^13 - 16*q^14 - 18*q^15 - 34*q^16 + O(q^17),
q^9 - 8*q^13 - 2*q^14 - 3*q^15 + 16*q^16 + O(q^17),
q^10 - 5*q^13 - 2*q^14 - 6*q^15 + 14*q^16 + O(q^17),
q^11 + 12*q^13 + 12*q^14 + 12*q^15 + 12*q^16 + O(q^17),
q^12 - 2*q^13 - q^14 + 2*q^15 + q^16 + O(q^17)

Hecke Algebras

Hecke Operators: Let p be a prime.

T_{p} (_{n≥1}^{∑} a_{n} · q^{n}) = _{n≥1}^{∑} a_{np} · q^{n} + p^{k−1} _{n≥1}^{∑} a_{n} · q^{np}

(If p | N, drop the second summand.) This preserves S_{k}(N), so defines a linear map

T_{p} : S_{k}(N) → S_{k}(N).

Similar definition of T_{n} for any integer n. Hecke Algebra: A commutative ring:

T_{k}(N) = Z[T_{1}, T_{2}, T_{3}, T_{4}, T_{5}, …] ⊂ End_{C}(S_{k}(N))

Computing Hecke Algebras

Fact: T_{k}(N) = Z[T_{1}, T_{2}, T_{3}, T_{4}, T_{5}, …] is free as a Z-module of rank equal to dim S_{k}(N).

Sturm Bound: T_{k}(N) is generated as a Z-module by T_{1}, T_{2}, …, T_{b}, where

b = ⌈\frac{k}{12} · N · _{p|N}^{∏} (1 + \frac{1}{p}) ⌉.

Example: For N = 37 and k = 2, the bound is 7. In fact,

T_{2}(37) has Z-basis T_{1} = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) and T_{2} = (\begin{matrix}−2 & 0 \\ 1 & 0 \end{matrix}).

There are several other T_{k}(N)-modules isomorphic to S_{2}(N), and I use these instead to compute T_{k}(N) as a ring.

Discriminants

The discriminant of T_{k}(N) is an integer. It measures ramification, or what’s the same, congruences between simultaneous eigenvectors for T_{k}(N), hence is related to the modular degree. Discriminant:

Disc (T_{k}(N)) = Det (Tr (t_{i} · t_{j})),

where t_{1}, …, t_{n} are a basis for T_{k}(N) as a free Z-module. Examples:

Disc (T_{2}(37)) = Det (\begin{matrix}2 & −2 \\ −2 & 4 \end{matrix}) = 4

Disc (T14 (11)) = 2^{46} · 3^{14} · 5^{2} · 11^{42} · 79 · 241 · 1163 · 40163 · 901181111 · 47552569849·124180041087631·205629726345973.

Ribet’s Question

I became interested in computing with modular forms when I was a grad student and Ken Ribet started asking:

Question: (Ribet, 1997) Is there a prime p so that p | Disc (T_{2}(p))?

Ribet proved a theorem about X_{0}(p)∩J_{0}(p)_{tor} under the hypothesis that p ∤ Disc (T_{2}(p)), and wanted to know how restrictive his hypothesis was. Note: When k > 2, usually p | Disc (T_{k}(p)).

Computations

Using a script of Joe Wetherell, I set up a computation on my laptop and found exactly one example in which p | Disc (T_{2}(p)). It was p = 389, now my favorite number.

Last year I checked that for p < 50000 there are no other examples in which p | Disc (T_{2}(p)). For this I used the Mestre method of graphs, which involves computing with the free abelian group on the supersingular j-invariants in F_{p^{2}} of elliptic curves.

Index in the Normalization

Let T̃_{k}(p) be the normalization of T_{k}(p). Since T_{k}(p) is an order in a product of number fields, T̃_{k}(p) is the product of the rings of integers of those number fields.

It turned out that Ribet could prove his theorem under the weaker hypothesis that p ∤ [T̃_{2}(p) : T_{2}(p)]. I was unable to find a counterexample to this divisibility. (Note: Matt Baker’s Ph.D. was a complete proof of the result Ribet was trying to prove, but used different methods.)

[[Picture on blackboard of Spec (T_{k}(p))]]

Conjecture 2

Conjecture 2. (–). If T_{2}(p) is the Hecke algebra associated to S_{2}(Γ_{0}(p)), then p does not divide the index of T_{2}(p) in its normalization.

The primes that divide [T̃_{2}(p) : T_{2}(p)] are called congruence primes. They are the primes of congruence between non-Gal(\overline{Q}/Q)- conjugate eigenvectors for T_{2}(p). Using this observation and another theorem of Ribet (and Wiles’s theorem), we see that Conjecture 2 implies that p does not divide the modular degree of any elliptic curve of conductor p. This is why Conjecture 2 implies Conjecture 1.

But is there any reason to believe Conjecture 2, beyond knowing that it is true for p < 50000?

Example of Weight k = 14

Totally stuck on Conjecture 2, so look at weight k > 2. We have

Disc (T14 (11)) = 2^{46} · 3^{14} · 5^{2} · 11^{42} · 79 · 241 · 1163 · 40163 · 901181111 ·

47552569849·124180041087631·205629726345973.

Notice the large power of 11. Upon computing the p-maximal order in T_{14} (11) ⊗_{Z} Q, we find that 11 ∤ Disc (T̃_{14} (11)), so all the 11 is in the index of T_{14} (11) in T̃_{14} (11). Thus

ord_{11} ([T̃_{14} (11) : T_{14} (11)]) = 21.

Data for k = 4

For inspiration, consider weight > 2.

Each row contains pairs p and ord_{p}(Disc (T_{4}(p))).

2  3  5  7  11  13  17  19  23  29  31  37  41  43  47  53  59
0  0  0  0  0  2  2  2  2  4  4  6  6  6  6  8  8
61  67  71  73  79  83  89  97  101 103 107 109 113 127 131 137 139
10  10  10  12  12  12  14  16  16  16  16  18  18  20  20  22  24
149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233
24 24 26 26 26 28 28 30 30 32 32 32 34 36 36 38 38
239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337
38 40 40 42 42 44 44 46 46 46 48 50 50 52 52 54 56
347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439
56 58 58 58 60 62 62 62 65 66 66 68 68 70 70 72 72
443 449 457 461 463 467 479 487 491 499
72 74 76 76 76 76 78 80 80 82

A Pattern?

F. Calegari (during a talk I gave): There is almost a pattern!!! Frank, Romyar Sharifi and I computed 2 · [T̃_{4}(p) : T_{4}(p)] and obtained the numbers as in the table, except for p = 389 (which gives 64) and 139 (which gives 22). We also considered many other examples... and found a pattern!

Conjecture 3

In all cases, we found the following amazing pattern: Conjecture 3. Suppose p ≥ k − 1. Then

ord_{p}([T̃_{k}(p) : T_{k}(p)]) = ⌊\frac{p}{12} ⌋ · \binom{k/2}{2} + a(p, k),

where

a(p, k) = \begin{cases}0 & if p ≡ 1 (mod 12), \\ 3 · \binom{⌈\frac{k}{6}⌉}{2} & if p ≡ 5 (mod 12), \\ 2 · \binom{⌈\frac{k}{4}⌉}{2} & if p ≡ 7 (mod 12), \\ a(5, k) + a(7, k) & if p ≡ 11 (mod 12). \end{cases}

Warning

The conjecture is false without the constraint that p ≥ k − 1. For example, if p = 5 and k = 12, then the conjecture predicts that the index is 0+3·1 = 3, but in fact ord_{p}([T̃_{k}(p) : T_{k}(p)]) = 5. In our data when k > p + 1, then the conjectural ord_{p} is often less than the actual ord_{p}.

Summary

For many years I had no idea whether there should or shouldn’t be mod p congruence between nonconjugate eigenforms. (I.e., whether p divides modular degrees at prime level.) By considering weight k ≥ 4, and computing examples, a simple conjectural formula emerged. When specialized to weight 2 this formula is the conjecture that there are no mod p congruences.

Future Direction. Explain why there are so many mod p congruences at level p, when k ≥ 4. See paper for a strategy.

Connection with Vandiver’s Conjecture? Investigate the connection between Conjecture 1 and Flach’s results on modular degrees annihilating Selmer groups.

This Concludes ANTS VI: THANKS!

Many thanks to the organizers (Sands, Kelly, Buell): , , and Duncan Buell