Because this equality is still conjectural, we will cal the former quantity -- then number of derivitaves of $L_E(s)$ that vanish at $s=1$ - the

*analytic rank*of $E$. The topic of this post is to address the question: given an elliptic curve $E$, how do we go about computing its analytic rank?

Before we can hope to answer this question, we need to know how to evaluate $L_E(s)$ itself for any given $s$. In the previous post I gave both the Euler product and Dirichlet series definitions for $L_E(s)$; to jog your memory, here's the Euler product of $L_E(s)$:

$$ L_E(s) = \prod_p \left(1-a_p p^{-s} + \epsilon_p p^{-2s}\right)^{-1}, $$

where the product runs over all prime numbers, $a_p = p+1-\#\{E(\mathbb{F}_p)\}$, and $\epsilon_p = 0$ if $p$ divides the conductor of $E$ and $1$ otherwise. The Dirichlet series is $L_E(s) = \sum_{n}a_n n^{-s}$, which is precisely what you get when you multiply out the Euler product. Note that we are suppresing the dependence on $E$ in both the $a_n$ and $\epsilon_p$ constants.

However, both the Euler product and Dirichlet series representations of $L_E(s)$ will only converge absolutely when the real part of $s$ exceeds $\frac{3}{2}$. Although the Sato-Tate Conjecture (now a theorem, but the name has stuck) implies that the expansions will in fact converge conditionally for $\Re{s}>\frac{1}{2}$, the convergence is so slow that attempting to evaluate $L_E(s)$ near the central point by multiplying or summing enough terms is

*horribly*inefficient. As such, we need a better way to evaluate $L_E(s)$ - and thankfully, such better ways do indeed exit.

Remember how I mentioned in the previous post that $L$-functions obey a very nice symmetry condition? Well, here's that condition exactly: first, we need to define something called the

*completed $L$-function*$\Lambda_E(s)$. This is just $L_E(s)$ multiplied by some extra factors. Specifically,

$$\Lambda_E(s) = N^{\frac{s}{2}}(2\pi)^{-s}\Gamma(s) L_E(s), $$

where $N$ is the conductor of $E$ and $\Gamma(s)$ is the usual Gamma function on $\mathbb{C}$ (the one that gives you $(n-1)!$ when you evaluate it at the positive integer $s=n$).

We can show that $\Lambda_E(s)$ is entire on $\mathbb{C}$; that is, it doesn't have any poles. Moreover, $\Lambda_E(s)$ obeys the glorious symmetry property

$$ \Lambda_E(s) = w_E \Lambda_E(2-s), $$

where $w_E$ is either $1$ or $-1$ and depends on $E$. This is called the

*functional equation*for the $L$-function attached to $E$.

Another way to put it is that shifted completed $L$-function $\Lambda_E(1+s)$ is either an even or an odd function of $s$. Because the factors $N^{\frac{s}{2}}$, $(2\pi)^{-s}$ and $\Gamma(s)$ are all readily computable, this allows us to determine the value of $L_E(s)$ when the real part of $s$ is less than $\frac{1}{2}$.

What's left, then, is to figure out how to efficinetly evaluate $L_E(s)$ in the strip $\frac{1}{2} \le \Re(s) \le \frac{3}{2}$. This is called the

*critical strip*for the $L$-function, and it is here that the behaviour of the function is most interesting.

[Aside: we can, for example, show that $\Lambda_E(1+s)$ is never zero outside of the critical strip. The Generalized Riemann Hypothesis in fact asserts that elliptic curve $L$-functions are only ever zero along the exact center of this strip, the

*critical line*$\Re(s)=1$. We'll get back to the zeros of elliptic curve $L$-functions in a later post.]

To evaluate $\Lambda_E(s)$ in the critical strip, we make use of the modularity of $E$. The modularity theorem states that elliptic curves are

*modular*: for every elliptic curve $E$ over the rationals there exists a weight 2 cuspidal eigenform $f_E$ of level $N$ (where $N$ is precisely the conductor of $E$), such that the $a_n$ of $E$ as defined previously equal the Fourier coefficients of $f_E$. If you haven't studied modular forms before, the crucial piece of knowledge is that there is a natural procedure for constructing $L$-functions from modular forms in such a way that the definition makes sense for all complex inputs $s$, and that the $L$-function attached to the cusp form $f_E$ will exactly equal the elliptic curve $L$-function $L_E(s)$. This is in fact how we show that elliptic curve $L$-functions can be analytically continued to all of $\mathbb{C}$.

The take-away is that via modularity there is an infinite sum representation for $L_E(s)$ which converges absolutely for all $s$. Here it is: define the auxiliary function

$$\lambda_E(s) = \left(\frac{\sqrt{N}}{2\pi}\right)^{s} \sum_{n=1}^\infty a_n n^{-s}\Gamma \left(s,\frac{2\pi}{\sqrt{N}}\cdot n\right), $$

where all the quantities are as defined previously, and $\Gamma(s,x)$ is the upper incomplete Gamma function (note that since $\Gamma(s,x)$ goes to zero exponentially as $x$ goes to infinity, this sum will converge absolutely for any $s$, with rate of convergence scaling with $\sqrt{N}$). Then we have

$$ \Lambda_E(s) = \lambda_E(s) + w_E \lambda_E(2-s). $$

Because we know how to compute incomplete Gamma functions, this gives us a way to evaluate $\Lambda_E(s)$, and thus $L_E(s)$, in the critical strip. The upside with this formula and variations thereof is that it works for any value of $s$ you stick in - including values near $s=1$. Similar formulae exist for the derivatives of $L_E(s)$, so we can in theory compute $L_E^{(n)}(1)$, the $n$th derivative of $L_E(s)$ at the central point, to any degree of accuracy for any $n$.

Thus if we want to compute a curve's analytic rank, what's stopping us from just evaluating successive derivatives of $L_E(s)$ at $s=1$ until we hit one which is not zero?

Two reasons. The first is that there's no way around the fact that you need about $\sqrt{N}$ terms to compute $L_E(s)$ or its derivatives to decent precision. If the conductor of the curve is too big, as is often the case, it takes an inordinate amount of time to simply evaluate the $L$-function near the central point. This makes direct evaluation impractical for all but the smallest-conductor curves -- and for those curves we can usually compute rank via other methods anyway.

The second reason is a more subtle one: how do you tell numerically if the $n$th derivative of $L_E(s)$ at $s=1$ is zero? If you think about it, it's easy to answer this question in one direction only: if you evaluate $L_E^{(n)}(1)$ to some precision and get digits that aren't all zero, then (assuming your code is correct) the $n$th derivative of $L_E(s)$ does not vanish. However, no matter how many digits of precision we compute getting all zeros, the possibility will always remain that the next digit along might not be zero.

In general, there is no numerical way to determine if the value of a black-box complex-valued function at a point is zero, or just really really close to zero. This is why, if you look in the literature, you'll find "methods to estimate analytic rank", but never to compute the quantity exactly. It's impossible to do without extra knowledge. Specifically, we'd need to have some sort of a computable lower bound on the size of the derivatives of $L_E(s)$ at the central. Unfortunately, no such theorems currently exist, so we're stuck with estimating analytic rank for now.

Thankfully, the $\sqrt{N}$-dependence issue is more hopeful. The next post will detail a method that provides good estimates for the analytic rank that scales much more slowly with the curve's conductor.

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