## Thursday, June 26, 2014

### Fun with elliptic curve L-function explicit formulae

Although I gave a brief exposition of the "baseline" explicit formula for the elliptic curve $L$-functions in a previous post, I wanted to spend some more time showing how one can use this formula to prove some useful statements about the elliptic curves and their $L$-functions. Some of these are already implemented in some way in the code I am building for my GSoC project, and I hope to include more as time goes on.

First up, let's fix some notation. For the duration of this post we fix an elliptic curve $E$ over the rationals with conductor $N$. Let $L_E(s)$ be its associated $L$-function, and let $\Lambda_E(s)$ be the completed $L$-function thereof, i.e.
$\Lambda_E(s) = N^{s/2}(2\pi)^s\Gamma(s)L_E(s)$, where $\Gamma$ is the usual Gamma function on $\mathbb{C}$. If you want to know more about how $E$, $L_E$ and $\Lambda_E$ are defined and how we compute with them, some of my previous posts (here and here) go into their exposition in more detail.

Let's recap briefly how we derived the baseline explicit formula for the $L$-function of $E$ (see this post for more background thereon). Taking logarithmic derivatives of the above formula for $\Lambda_E(s)$ and shifting to the left by one gives us the following equality:
$$\frac{\Lambda_E^{\prime}}{\Lambda_E}(1+s) = \log\left(\frac{\sqrt{N}}{2\pi}\right) + \frac{\Gamma^{\prime}}{\Gamma}(1+s) + \frac{L_E^{\prime}}{L_E}(1+s).$$
Nothing magic here yet. However, $\frac{\Lambda_E^{\prime}}{\Lambda_E}$, $\frac{\Gamma^{\prime}}{\Gamma}$ and $\frac{L_E^{\prime}}{L_E}$ all have particularly nice series expansions about the central point. We have:

• $\frac{\Lambda_E^{\prime}}{\Lambda_E}(1+s) = \sum_{\gamma} \frac{s}{s^2+\gamma^2}$, where $\gamma$ ranges over the imaginary parts of the zeros of $L_E$ on the critical line; this converges for any $s$ not equal to a zero of $L_E$.
• $\frac{\Gamma^{\prime}}{\Gamma}(1+s) = -\eta + \sum_{k=1}^{\infty} \frac{s}{k(k+s)}$, where $\eta$ is the Euler-Mascheroni constant $=0.5772156649\ldots$ (this constant is usually denoted by the symbol $\gamma$ - but we'll be using that symbol for something else soon enough!); this sum converges for all $s$ not equal to a negative integer.
• $\frac{L_E^{\prime}}{L_E}(1+s) = \sum_{n=1}^{\infty} c_n n^{-s}$; this only converges absolutely in the right half plane $\Re(s)>\frac{1}{2}$.

Here
$$c_n = \begin{cases} \left[p^m+1-\#E(\mathbb{F}_{p^m})\right]\frac{\log p}{p^m}, & n = p^m \mbox{ is a perfect prime power,} \\ 0 & \mbox{otherwise.}\end{cases}$$

Assembling these equalities gives us the aforementioned explicit formula:
$$\sum_{\gamma} \frac{s}{s^2+\gamma^2} = \left[-\eta+\log\left(\frac{\sqrt{N}}{2\pi}\right)\right] + \sum_{k=1}^{\infty} \frac{s}{k(k+s)} + \sum_n c_n n^{-s}$$
which holds for any $s$ where all three series converge. It is this formula which we will use repeatedly to plumb the nature of $E$.

For ease of exposition we'll denote the term in the square brackets $C_0$. It pitches up a lot in the math below, and it's a pain to keep writing out!

Some things to note:

• $C_0 = -\eta+\log\left(\frac{\sqrt{N}}{2\pi}\right)$ is easily computable (assuming you know $N$). Importantly, this constant depends only on the conductor of $E$; it contains no other arithmetic information about the curve, nor does it depend in any way on the complex input $s$.
• The sum $\sum_{k=1}^{\infty} \frac{s}{k(k+s)}$ doesn't depend on the curve at all. As such, when it comes to computing values associated to this sum we can just hard-code the computation before we even get to working with the curve itself.
• The coefficients $c_n$ can computed by counting the number of points on $E$ over finite fields up to some bound. This is quick to do for any particular prime power.

Good news: the code I've written can compute all the above values quickly and efficiently:

sage: E = EllipticCurve([-12,29])
sage: Z = LFunctionZeroSum(E)
sage: N = E.conductor()
sage: print(Z.C0(),RDF(-euler_gamma+log(sqrt(N)/(2*pi))))
(2.0131665172, 2.0131665172)
sage: print(Z.digamma(3.5),RDF(psi(3.5)))
(1.10315664065, 1.10315664065)
sage: Z.digamma(3.5,include_constant_term=False)
1.68037230555
sage: Z.cn(389)
-0.183966457901
sage: Z.cn(next_prime(1e9))
0.000368045198812
sage: timeit('Z.cn(next_prime(1e9))')
625 loops, best of 3: 238 µs per loop

So by computing values on the right we can compute the sum on the left - without having to know the exact locations of the zeros $\gamma$, which in general is hard to compute.

Now that we have this formula in the bag, let's put it to work.

### NAÏVE RANK ESTIMATION

If we multiply the sum over the zeros by $s$ and letting $\Delta = 1/s$, we get
$$\sum_{\gamma} \frac{\Delta^{-2}}{\Delta^{-2}+\gamma^2} = \sum_{\gamma} \frac{1}{1+(\Delta\gamma)^2},$$
Note that for large values of $\Delta$, $\frac{1}{1+(\Delta\gamma)^2}$ is small but strictly positive for all nonzero $\gamma$, and 1 for the central zeros, which have $\gamma=0$. Thus the value of the sum when $\Delta$ is large gives a close upper bound on the analytic rank $r$ of $E$. That is, we can bound the rank of $E$ from above by choosing a suitably large value of $\Delta$ and computing the quantity on the right in the inequality below:
$$r < \sum_{\gamma} \frac{1}{1+(\Delta\gamma)^2} = \frac{1}{\Delta}\left[C_0 + \sum_{k=1}^{\infty} \frac{1}{k(1+\Delta k)} + \sum_n c_n n^{-1/\Delta}\right].$$
Great! Right? Wrong. In practice this approach is not a very good one. The reason is the infinite sum over $n$ only converges absolutely for $\Delta<2$, and for Delta values as small as this, the obtained bound won't be very good. A value of $\Delta=2$, for example, gives us the zero sum $\sum_{\gamma} \frac{1}{1+(2\gamma)^2}$. If a curve's $L$-function has a zero with imaginary part about 0.5, for example - as many $L$-functions do - then such a zero will contribute 0.5 to the sum. And because zeros come in symmetric pairs, the sum value will then be at least a whole unit larger than the actual analytic rank. In general, for $\Delta<2$ the computed sum can be quite a bit bigger than the actual analytic rank of $E$.

Moreover, even though the Generalized Riemann Hypothesis predicts that the sum $\sum_n c_n n^{-1/\Delta}$ does indeed converge for any positive value of $\Delta$, in practice the convergence is so slow that we end up needing to compute inordinate amounts of the $c_n$ in order to hope to get a good approximation of the sum. So no dice; this approach is just too inefficient to be practical.