Tuesday, July 8, 2014

The Birch and Swinnerton-Dyer Conjecture and computing analytic rank

Let $E$ be an elliptic curve with $L$-function $L_E(s)$. Recall that Google Summer of Code project is to implement in Sage a method that allows us to compute $\sum_{\gamma} f(\Delta \gamma)$, where $\gamma$ ranges over the imaginary parts of the nontrivial zeros of $L_E$, $\Delta$ is a given positive parameter, and $f(x)$ is a specified symmetric continuous integrable function that is 1 at the origin. The value of this sum then bounds the analytic rank of $E$ - the number of zeros at the central point - from above, since we are summing $1$ with multipliticy $r_{an}$ in the sum, along with some other nonzero positive terms (that are hopefully small). See this post for more info on the method.

One immediate glaring issue here is that zeros that are close to the critical point will contribute values that are close to 1 in the sum, so the curve will then appear to have larger analytic rank than it actually does. An obvious question, then, is to ask: how close can the noncentral zeros get to the central point? Is there some way to show that they cannot be too close? If so, then we could figure out just how large of a $\Delta$ we would need to use in order to get a rank bound that is actually tight.

The rank zero curve $y^2 = x^3 -  x^3 - 7460362000712 x - 7842981500851012704$ has an extremely low-lying zero at $\gamma = 0.0256$ (and thus another one at $-0.0256$; as a result the zero sum looks like it's converging towards a value of 2 instead of the actual analytic rank of zero. In order to actually get a sum value out that's less than one we would have to use a $\Delta$ value of about 20; this is far beyond what's feasible due to the exponential dependence of the zero sum method on $\Delta$.

The good news is that there is hope in this regard; the nature of low-lying zeros for elliptic curve $L$-functions is actually the topic of my PhD dissertation (which I'm still working on, so I can't provide a link thereto just yet!). In order to understand how close the lowest zero can get to the central point we will need to talk a bit about the BSD Conjecture.

The Birch and Swinnerton-Dyer Conjecture is one of the two Clay Math Millenium Problems related to $L$-functions. The conjecture is comprised of two parts; the first part I mentioned briefly in this previous post. However, we can use the second part to gain insight into how good our zero sum based estimate for analytic rank will be.

Even though I've stated the first part of the BSD Conjecture before, for completeness I'll reproduce the full conjecture here. Let $E$ be an elliptic curve defined over the rational numbers, e.g. a curve represented by the equation $y^2 = x^3 + Ax + B$ for some integers $A$ and $B$ such that $4A^3+27B^2 \ne 0$. Let $E(\mathbb{Q})$ be the group of rational points on the elliptic curve, and let $r_{al}$ be the algebraic rank of $E(\mathbb{Q})$. Let $L_E(s)$ be the $L$-function attached to $E$, and let $L_E(1+s) = s^{r_{an}}\left[a_0 + a_1 s + \ldots\right]$ be the Taylor expansion of $L_E(s)$ about $s=1$ such that the leading coefficient $a_0$ is nonzero; $r_{an}$ is called the analytic rank of $E$ (see here for more details on all of the above objects). The first part of the BSD conjecture asserts that $r_{al}=r_{an}$; that is, the order of vanishing of the $L$-function about the central point is exactly equal to the number of free generators in the group of rational points on $E$.

The second part of the conjecture asserts that we actually know the exact value of that leading coefficient $a_0$ in terms of other invariants of $E$. Specifically:
$$ a_0 = \frac{\Omega_E\cdot\text{Reg}_E\cdot\prod_p c_p\cdot\#\text{Ш}(E/\mathbb{Q})}{(\#E_{\text{Tor}}(\mathbb{Q}))^2}. $$

Fear not if you have no idea what any of these quantities are. They are all things that we know how to compute - or at least estimate in size. I provide below brief descriptions of each of these quantities; however, feel free to skip this part. It suffices to know that we have a formula for computing the exact value of that leading coefficient $a_0$.
  1. $\#E_{\text{Tor}}(\mathbb{Q})$ is the number of rational torsion points on $E$. Remember that the solutions $(x,y)$ to the elliptic curve equation $y^2 = x^3 + Ax+B$, where $x$ and $y$ are both rational numbers, form a group. Recall also that that the group of rational points $E(\mathbb{Q})$ may be finite or infinite, depending on whether the group has algebraic rank zero, or greater than zero. However, it turns out that there are only ever finitely many torsion points - those which can be added to themselves some finite number of times to get the group identity element. These points of finite order form a subgroup, denoted $E_{\text{Tor}}(\mathbb{Q})$, and the quantity in question is just the size of this finite group (squared in the formula). In fact, it's been proven that the size of $E_{\text{Tor}}(\mathbb{Q})$ is at most 16.
  2. $\Omega_E$ is the real period of $E$. This is perhaps a bit more tricky to define, but it essentially is a number that measures the size of the set of real points of $E$. If you plot the graph of the equation representing $E: y^2 = x^3 + Ax + B$ on the cartesian plane, you get something that looks like one of the following:

    The plots of the real solutions to four elliptic curves, and their associated real periods.

    There is a way to assign an intrinsic "size" to these graphs, which we denote the real period $\Omega_E$. The technical definition is that $\Omega_E$ is equal to the integral of the absolute value of the differential $\omega = \frac{dx}{2y}$ along the part of the real graph of $E$ that is connected to infinity (that or twice that depending on whether the cubic equation $x^3 + Ax + B$ has one or three real roots respectively).
  3. $\text{Reg}_E$ is the regulator of $E$. This is a number that measures the "density" of rational points on $E$. Recall that $E(\mathbb{Q}) \approx T \times \mathbb{Z}^{r_{an}}$, i.e there free part of $E(\mathbb{Q})$ is isomorphic to $r_{an}$ copies of the integers. There is a canonical way to embed the free part of $E(\mathbb{Q})$ in $\mathbb{R}^{r_{an}}$ as a lattice; the regulator $\text{Reg}_E$ is the volume of the fundamental domain of this lattice. The thing to take away here is that elliptic curves with small regulators have lots of rational points whose coefficients have small numerators and denominators, while curves with large regulators have few such points.
  4. $\prod_p c_p$ is the Tamagawa product for $E$. For each prime $p$, one can consider the points on $E$ over the $p$-adic numbers $\mathbb{Q}_p$. The Tamagawa number $c_p$ is the ratio of the size of the full group of $p$-adic points on $E$ to the subgroup of $p$-adic points that are connected to the identity. This is always a positive integer, and crucially, in all but a finite number of cases the ratio is 1. Thus we can consider the product of the $c_p$ as we range over all prime numbers, and this is precisely the definition of the Tamagawa product.
  5. $\#\text{Ш}(E/\mathbb{Q})$ is the order of the Tate-Shafarevich group of $E$ over the rational numbers. The Tate-Shafarevich group of $E$ is probably the most mysterious part of the BSD formula; it is defined as the subgroup of the Weil–Châtelet group $H^1(G_{\mathbb{Q}},E)$ that becomes trivial when passing to any completion of $\mathbb{Q}$. If you're like me then this definition will be completely opaque; however, we can think of $\text{Ш}(E/\mathbb{Q})$ as measuring how much $E$ violates the local-global principle: that one should be able to find rational solutions to an algebraic equation by finding solutions modulo a prime number $p$ for each $p$, and then piecing this information together with the Chinese Remainder Theorem to get a rational solution. Curves with nontrivial $\text{Ш}$ have homogeneous spaces that have solutions modulo $p$ for all $p$, but no rational points. The main thing here is that $\text{Ш}$ is conjectured to be finite, in which case $\#\text{Ш}(E/\mathbb{Q})$ is just a positive integer (in fact, it can be shown for elliptic curves that if $\text{Ш}$ is indeed finite, then its size is a perfect square).
Why is the BSD Conjecture relevant to rank estimation? Because it helps us overcome the crucial obstacle to computing analytic rank exactly: without extra knowledge, it's impossible to decide using numerical techniques whether the $n$th derivative of the $L$-function at the central point is exactly zero, or just so small that it looks like it is zero to the precision that we're using. If we can use the BSD formula to show a priori that $a_0$ must be at least $M_E$ in magnitude, where $M_E$ is some number that depends only on some easily computable data attached to the elliptic curve $E$, then all we need to do is evaluate successive derivatives of $L_E$ at the central point to enough precision to decide if that derivative is less than $M_E$ or not; this is readily doable on a finite precision machine. Keep going until we hit a derivative which is then definitely greater than $M_E$ in magnitude, at which we can halt and declare that the analytic rank is precisely the order of that derivative.

In the context of the explicit formula-based zero sum rank estimation method implemented in our GSoC project, the size of the leading coefficient also controls how far close the lowest noncentral zero can be from the central point. Specifically, we have the folling result: Let $\gamma_0$ be the lowest-lying noncentral zero for $L_E$ (i.e. the zero closest to the central point that is not actually at the central point); let $\Lambda_E(s) = N^{s/2}(2\pi)^s\Gamma(s)L_E(s)$ is the completed $L$-function for $E$, and let $\Lambda_E(1+s) = s^{r_{an}}\left[b_0 + b_1 s + b_2 s^2 \ldots\right]$ be the Taylor expansion of $\Lambda_E$ about the central point. Then:
$$ \gamma_0 > \sqrt{\frac{b_0}{b_2}}. $$
Thankfully, $b_0$ is easily relatable back to the constant $a_0$, and we have known computable upper bounds on the magnitude of $b_2$, so knowing how small $a_0$ is tells us how close the lowest zero can be to the central point. Thus bounding $a_0$ from below in terms of some function of $E$ tells us how large of a $\Delta$ value we need to use in our zero sum, and thus how long the computation will take as a function of $E$.

If this perhaps sounds a bit handwavy at this point it's because this is all still a work in progress, to be published as part of my dissertation. Nevertheless, the bottom line is that bounding the leading $L$-series coefficient from below gives us a way to directly analyze the computational complexity of analytic rank methods. 

I hope to go into more detail in a future post about what we can do to bound from below the leading coefficient of $L_E$ at the central point, and why this is a far from straightforward endeavour.


  1. I am trying to understand the genesis of your rank 0 example with a low height zero. It seems to me that you want small conductor as the lowest lying zero is expected at 2*Pi/log(N), versus the size of the real period for the latter dominates the BSD-quotient and is something like Delta^(-1/12). In other words, you want good ABC-examples with (log Delta/log N) as large as possible? Is the example you chose the best one (N=256944) in Cremona's database?

    In other words, taking Sha=Tors=1, the L-value is about 1/omega, though in these ABC examples you should account Tamagawa numbers too (here the product is 2). This small size of L(E,1) is then redolent of low-lying zeros by basic calculus, the second derivative that you display measures this more exactly but is of a generic nontrivial size (a power of log(N)?).

    Actually, maybe asymptotically the 2*Pi/log(N) is irrelevant, and you only want examples where 1/omega is small, giving a small L-value and then presumably close zeros. OTOH, to be able to compute a numerical example in the way you did, you need the conductor small enough to compute the zeros by a sqrt(N) method (Rubinstein's L package).

    With point #3, I agree the regulator plays some role here (as the covolume of the lattice), but the rank also matters. Asymptotically it is more important, thanks to the ellipsoid asymptotic h^(r/2)/sqrt(Reg). So it is not quite correct to say that curves with smaller regulators necessarily have more small points.

    P.S. Don't call your scaling or truncation variable Delta, that should be the discriminant!

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  3. Hi CatacycleUmberto

    Sorry, attempt number one got your name wrong. Quoth:

    Yes, the curve with conductor 256944 is simply the one in the Cremona database with lowest nontrivial zero. Not much more thought than that went into its selection; I was merely looking for an example that demonstrated how this rank estimation method could fail.

    Alas, the choice of Delta as the scaling parameter name is not my own; I'm following the notation in Bober's ANTS paper (http://msp.org/obs/2013/1-1/obs-v1-n1-p07-s.pdf). I use the parameter $t$ in the nuts and bolts of my code (where $\Delta = 2\pi t$, as then the bound up to which one must compute logarithmic derivative coefficients is exactly $e^t$), though this parameter name again was chosen somewhat arbitrarily.

    And yes, you are correct: the exact relationship between regulator and the number of 'small' points is more refined than I state in my post. However, one could make the argument that (hypothetically) a negligible proportion of curves have rank > 1, so the ellipsoid asymptotic doesn't come into play most of the time...

    - Simon