$$ r < \sum_{\gamma} \text{sinc}^2(\Delta\gamma) = \frac{1}{\pi\Delta} \left[ C_0 + \frac{1}{2\pi\Delta}\left(\frac{\pi^2}{6}-\text{Li}_2\left(e^{-2\pi\Delta}\right) \right) + \sum_{n=1}^{e^{2\pi\Delta}} c_n\left(1 - \frac{\log n}{2\pi\Delta}\right) \right] $$

where

- $\gamma$ ranges over the nontrivial zeros of the $L$-function attached to $E$
- $\Delta$ is a positive parameter
- $C_0 = -\eta + \log\left(\frac{\sqrt{N}}{2\pi}\right)$; $\eta$ is the Euler-Mascheroni constant $=0.5772\ldots$ and $N$ is the conductor of $E$
- $\text{Li}_2(x)$ is the dilogarithm function, defined by $\text{Li}_2(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$
- $c_n$ is the $n$th coefficient of the logarithmic derivative of the $L$-function of $E$.

The thing I want to look at in this post is that parameter $\Delta$. The larger you make it, the closer the sum on the left hand side over the zeros is to the analytic rank, so when trying to determine the rank of $E$ we want to pick as large a $\Delta$ as we can. However, the bigger this parameter is the more terms we have to compute in the sum over the $c_n$ on the right hand side; moreover the number of terms - and thus the total computation time - scales exponentially with $\Delta$. This severely constrains how big we can make $\Delta$; generally a value of $\Delta=2$ may take a second or two for a single curve on SageMathCloud, while $\Delta=3$ may take upwards of an hour. For the average rank project I'm working on I ran the code on 12 million curves using $\Delta=1.3$; the total computation time was about 4 days on SMC.

However, it should be clear that using too large a $\Delta$ is overkill: if you run the code on a curve with $\Delta=1$ and get a bound of zero out, you know that curve's rank exactly zero (since it's at most zero, and rank is a non-negative integer). Thus using larger $\Delta$ values on that curve will do nothing except provide you the same bound while taking much longer to do so.

This begs the question: just how big of a $\Delta$ value is good enough? Can we, given some data defining an elliptic curve, decide a priori what size $\Delta$ to use so that a) the computation returns a bound that is likely to be the true of the curve, and b) it will do so in as little time as possible?

The relevant invariant to look at here is conductor $N$ of the elliptic curve; go back to the formula above and you'll see that the zero sum includes a term which is $O\left(\frac{\log(N)}{2\pi\Delta}\right)$ (coming from the $\frac{1}{\pi \Delta} C_0$ term). This means that size of the returned estimate will scale with $\log(N)$: for a given $\Delta$, the bound returned on a curve with 10-digit conductor will be about double that which is returned for a 5-digit conductor curve, for example. However, we can compensate this by increasing $\Delta$ accordingly.

To put it all more concretely we can pose the following questions:

- Given an elliptic curve $E$ with conductor $N$, how large does $\Delta$ need to be in terms of $N$ so that the returned bound is
*guaranteed*to be the true analytic rank of the curve? - Given a conductor size $N$ and a proportionality constant $P \in [0,1]$, how big does $\Delta$ have to be in terms of $N$ and $P$ so that at least $P\cdot 100$ percent of all elliptic curves with conductor of size about $N$ will, when the rank estimation code is run on them with that $\Delta$ value, yield returned bounds that are equal to their true analytic rank?

[Note: in both of the above questions we are making the implicit assumption that the returned rank bound is monotonically decreasing for increasing $\Delta$. This is not necessarily the case: the function $y = \text{sinc}^2(x)$ is

*not*a decreasing function in $x$. However, in practice any upwards fluctuation we see in the zero sum is small enough to be eliminated when we take the floor function to get an integer rank bound.]### A $\Delta$ CHOICE GOOD ENOUGH FOR MOST CURVES

The first question is easier to phrase, but more difficult to answer, so we will defer it for now. To answer the second question, it is useful mention what we know about the distribution and density of nontrivial zeros of an elliptic curve $L$-function.

Using some complex analysis we can show that, for the $L$-function of an elliptic curve with conductor $N$, the expected number of zeros in the critical strip with imaginary part at most $T$, is $O(T\log N+ T\log T)$. That is, expected zero density has two distinct components: a part that scales linearly with log of the conductor, and a part that doesn't scale with conductor (but does scale slightly faster than linearly with how far out you go).

Consider the following: if we let

$$\Delta(N) = \frac{C_0}{\pi} = \frac{1}{\pi}\left(-\eta+\log\left(\frac{\sqrt{N}}{2\pi}\right)\right)$$

then the first term in the right hand side of the zero sum formula is precisely 1 - this is the term that comes from the $\log N$ part of the zero density. The next term - the one involving $\frac{\pi^2}{6}-\text{Li}_2\left(e^{-2\pi\Delta}\right)$ - is the term that comes from the part independent of $N$; because the right hand side is divided by $\Delta$ it therefore goes to zero as the curve's conductor increases. The third term contains the $c_n$ coefficients which (per Sato-Tate) will be positive half the time and negative half the time, so the entire sum could be positive or negative; we therefore expect its contribution to be small on average when we look at large number of elliptic curves.

It thus stands to reason that for this value of Delta, and when the conductor $N$ is sufficiently large, the zero sum will be about 1, plus or minus a smallish amount coming from the $c_n$ sum. This argument is by no means rigorous, but we might therefore expect the zero sum to be within 2 of the actual analytic rank most of the time. Couple that with knowledge of the root number and you get a rank upper bound out which is equal to the actual analytic rank in all but a few pathological cases.

Empirical evidence bears this out. I ran my rank estimation code with this choice of $\Delta$ scaling on the whole Cremona database, which contains all elliptic curves up to conductor 350000:

As you can see, the percentage of curves for which the rank bound is strictly greater than rank holds fairly constant at about 0.25%. That's minuscule: what this is saying is that if you type in a random Weierstrass equation, there is only about a 1 in 1000 chance that the rank bounding code with $\Delta = \frac{C_0}{\pi}$ will return a value that isn't the actual analytic rank. Moreover, the code runs

*considerably*faster than traditional analytic rank techniques, so if you wanted to, for example, compute the ranks of a database of millions of elliptic curves, this would be a good first port of call.
Of course, this $\Delta$ scaling approach is by no means problem-free. Some more detailed analysis will show that that as stated above, the runtime of the code will actually be $O(N)$ (omitting log factors), i.e. asymptotic scaling is actually

*worse*than traditional analytic rank methods, which rely on evaluating the $L$-function directly and thus are $O(\sqrt{N})$. It's just that with this code we have some very small constants sitting in front, so the crossover point is at large enough conductor values that neither method is feasible anyway.
This choice of $\Delta$ scaling works for conductor ranges up to about $10^9$; that corresponds to $\Delta \approx 2.5$, which will give you a total runtime of about 10-20 seconds for a single curve on SMC. Increase the conductor by a factor of 10 and your runtime will also go up tenfold.

For curves of larger conductor, instead of setting $\Delta = \frac{C_0}{\pi}$ we can choose to set $\Delta$ to be $\alpha\cdot \frac{C_0}{\pi}$ for any $\alpha \in [0,1]$; the resulting asymptotic runtime will then be $O(N^{\alpha})$, at the expense of having a reduced proportion of elliptic curves where rank bound is equal to true rank.

### HOW LARGE DOES $\Delta$ HAVE TO BE TO GUARANTEE TRUE RANK?

When we use $\Delta = \frac{C_0}{\pi}$, the curves for which the returned rank estimate is strictly larger than the true rank are precisely those which have unusually low-lying zeros. For example, the rank 0 curve with Cremona label 256944c1, has a zero with imaginary part at 0.0256 (see here for a plot), compared to an expected value of 0.824. Using $\Delta = \frac{C_0}{\pi}$ on this curve means $\Delta \approx 1.214$; if we compute the corresponding zero sum with this value of $\Delta$ we get a value of 2.07803. The smallest value of $\Delta$ for which we get a zero sum value of less than 2 is empirically about 2.813; at this point taking the floor and invoking parity tells us that the curve's rank is zero.

The above example demonstrates that if we want to guarantee that the returned rank bound is the true analytic rank, we are forced to increase the size of $\Delta$ to something larger than $\frac{C_0}{\pi}$. Do we need to increase $\Delta$ by a fixed value independent of $N$? Do we need to increase it by some constant factor? Or does it need to scale faster than $\log N$? These are hard questions to answer; it comes down to determining how close to the central point the lowest nontrivial zero can be as a function of the conductor $N$ (or some other invariants of $E$), which in turn is intimately related to estimating the size of the leading coefficient of the $L$-series of $E$ at the central point. This is already the topic of a previous post: it is a question that I hope to make progress in answering in my PhD dissertation.

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