mixedmath

Explorations in math and programming
David Lowry-Duda



Suppose we have an elliptic curve $E$ over $\mathbb{Q}$ of conductor $N$. How many coefficients $a_p(E)$ does it take to uniquely identify the isogeny class of $E$? (Two elliptic curves over $\mathbb{Q}$ are isogenous if and only if they have the same $L$-function, so we cannot hope to distinguish isogenous elliptic curves by individual point counts).

This question is at the heart of the paper BCCHLLDNP25, part of a collaboration with Babei, Charton, Costa, Huang, Lee, Narayanan, and Pozdnyakov. In this paper we see if ML models can learn to predict later coefficients from collections of earlier coefficients. Successful predictions would indicate that there is enough structure in early coefficients to uniquely specify later coefficients — and therefore that the initial sets of coefficients uniquely specify the curve.

It would be desirable to directly try to test the classification problem of how many coefficients determine a curve, but it's challenging to formulate a concrete, testable experiment for this. (We return to this in the LMFDB Experiment section below).

Instead we test the presumably harder problem of reconstructing later coefficients from earlier coefficients. This is (probably) a much harder problem!

Extending lists of coefficients

How could one go about recognizing or reconstructing a curve from initial sets of coefficients?

Naive Approach

Suppose we are given $a_2, a_3, \ldots, a_p$ for some number of primes $p$. Equivalently, we are given the number of rational points for some curve $E$ over several finite fields $\mathbb{F}_p$.

How hard is it to directly find a curve with these points?

One approach is to consider it one prime at a time and then to combine the local results together with an application of the Chinese Remainder Theorem.1 1This is like Schoof's algorithm, but in reverse. For example, guess random curves $Y^2 = X^3 + aX + b$ for random $a$ and $b$ in $\mathbb{F}_p$.2 2I'm assuming implicitly here that we look at the short Weierstrass model. Thus one shouldn't look at $\mathbb{F}_2$ or $\mathbb{F}_3$ here.

I thought it would be fun to use this to find matching candidates for 11a3. And it was fun! But it also worked terribly.

Any initial sequence of $a_p$ coefficients comes from an elliptic curve. Indeed, this construction generates a list of those potential curves. All of these curves are reasonable, but this construction doesn't have any restrictions on the conductor of the generated curves. Thus one gets generates a list of elliptic curves with large conductor, usually. (It would be separately interesting to try to understand more about what random conductors appear, statistically. That's for another day).

Point counts reduce to $2$ candidate pairs over $\mathbb{F}_5$, $6$ over $\mathbb{F}_7$, $15$ over $\mathbb{F}_{13}$, $28$ over $\mathbb{F}_{17}$, and $36$ over $\mathbb{F}_{19}$. In total, this gives $181440$ candidate pairs $(a, b)$ mod $5 \cdot 7 \cdot 13 \cdot 13 \cdot 19$. For each of these residue classes, I tried the four smallest (in absolute value) lifts to $\mathbb{Z}$.

And yes, it worked!

But coincidentally it was approximately the $300{,}000$th constructed curve that I tried.3 3I suppose I was going to try $4 \cdot 181440$ in total, and this is approximately halfway through. Maybe that is to be expected? This was a lot of work!

Sturm's Bound

Concretely, we know that the first $\asymp N$ coefficients will uniquely determine (an isogeny class of) an elliptic curve of conductor $N$ because of the modularity theorem. There is a corresponding modular form of level $N$, and Sturm's bound shows that the first $\asymp N$ coefficients determine the form.

Thus given $\asymp N$ coefficients, one could then construct the entire space of weight $2$, level $N$ modular forms and pluck out the matching one. (This is also computationally demanding).

Approximate Functional Equation

The $L$-function of $E$ will satisfy a functional equation of the shape \begin{equation*} \Lambda(s) := N^{s/2} \Gamma_{\mathbb{C}}(s) L(s) = \pm \Lambda(2 - s), \end{equation*} implying an approximate functional equation that implies morally \begin{equation} \Lambda(s) \approx \sum_{n \geq 1} \frac{a_n(E)}{n^s} V_s(n/\sqrt{N}) \end{equation} for a weight function $V_s(y)$ that decays rapidly when $y > 1$. Stated differently, $\Lambda(s)$ is approximately completely determined by the first $\sqrt{N}$ coefficients. (There is a dual sum of the same length that I omit).

In principle, the coefficients can be retrieved from the analytic behavior of the $L$-function. In practice, this is possibly easier by using a variety of weight functions $V_s$ and combining them in clever ways — Farmer, Koutsoliotas, and Lemurell do this with harder approximate functional equations and less-well behaved objects in their paper on maass forms on GL3 and GL4, for instance.

I don't know of any implementation for the purpose of extending coefficient lists (though this would also be interesting and attainable!), but it is performable in principle. Thus the first $O(\sqrt{N})$ coefficients specify the form and "are enough" to reconstruct later coefficients.

Conjectural Results

If the Riemann Hypothesis is true for Artin $L$-functions, then we actually expect that the first $O(\log^2(N))$ coefficients are enough to uniquely specify the isogeny class. (See Bucur and Kedlaya's An application of the effective Sato-Tate conjecture, Corollary 4.8 for more on this).

This says nothing about how to use these to compute the remaining coefficients.

LMFDB Experiment

Let's look at the LMFDB's elliptic curves over $\mathbb{Q}$ up to conductor $500{,}000$ (where the table is complete). For every conductor $N$, we compute $B = B(N)$ such that if $E_1$ and $E_2$ are two elliptic curves of conductor $N$, then there exists some $p \leq B$ such that $a_p(E_1) \neq a_p(E_2)$.

Stated differently, we compute the number of $a_p$ coefficients are sufficient to distinguish among all the isogeny classes of elliptic curves of that conductor.

This is a very short data analysis problem with the database available.

# This line is made up, but is approximately right.
# (If you need help accessing LMFDB data, send me
#  or the LMFDB an email and we can help.)
from lmfdb import elliptic_curve_db

def identify_smallest(apcol, conductor, record):
    """
    -1 means unable to distinguish
    0 means only one isogeny class
    otherwise, return B
    """
    m = -1
    if len(apcol) == 1:
        record[conductor] = 0
        return
    for i in range(len(apcol)):
        for j in range(i + 1, len(apcol)):
            found = False
            for idx, (a, b) in enumerate(zip(apcol[i], apcol[j])):
                if a != b:
                    m = max(m, idx)
                    found = True
                    break
            if not found:
                record[conductor] = -1
                return
    record[conductor] = m

record = dict()
cur_conductor = 11
apcol = []
cursor = elliptic_curve_db.search(sort=['conductor'])

while True:
    ec = next(cursor)
    conductor = ec['conductor']
    if conductor != cur_conductor:
        identify_smallest(apcol, cur_conductor, record)
        cur_conductor = conductor
        apcol = []
    apcol.append(ec['aplist'])

Let's plot $B(N)$ and a reasonable curve. Below, I plot $B(N)$ and the curve $\log(N)$.

Looking a bit closer at the smaller range:

Maybe $\log(N)$ is about right?

Here is one heuristic explanation: we should expect $a_p(E)$ to be semi-circle-random between $-2 \sqrt{p}$ and $2 \sqrt{p}$. Assuming independence for simplicity, specifying $a_p(E)$ thus cuts out a region of density $1/4\sqrt{p}$ from the overall distribution.

There are $\ll N$ isogeny classes of elliptic curves of conductor $N$ (by modularity, these all correspond to a weight $2$ modular form of level $N$. The valence formula there gives the dimension of the space to be $\ll N$).

Heuristically, asking how many $a_p$ it takes to uniquely specify an isogeny class reduces to asking when \begin{equation*} \prod_{p \leq x} \frac{1}{4 \sqrt{p}} \asymp N. \end{equation*} Primorial asymptotics show this occurs approximately when $x \asymp \log N$. Thus maybe we should expect $\log(N)$ to be enough.

I've implicitly assumed that the distribution is uniform, and not merely Sato-Tate uniform. This is close enough to the truth to not affect the heuristic.

But what should we actually expect about the joint distributions across multiple coefficients?

Related Notes

These bounds are closely related to the Sturm Bound, Trace Bound, and Distinguishing $T_p$ bound for classical modular forms. (These are links to LMFDB knowls).

In particular, we look at Hecke kernels, distinguishing sets of Hecke operators $T_p$ acting on Galois orbits of newforms. The size of these kernels is often very small, perhaps much smaller than one might expect. On the other hand, the kernels in the LMFDB are constructed greedily in terms of the size of the prime, but the largest prime is sometimes larger than the Sturm bound.

This is to say: it's not quite the same... but it's closely related.


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