# Determining elliptic curve analytic rank even/odd

For an elliptic curve over Q that is defined with large coefficients, it can take mathematical software (such as Sage) a long to time calculate the analytic rank. However, it seems to quickly know if the rank is even or odd.

I would like to understand how they determine this so quickly.

This is hinted at in a PlanetMath article as the "root number" obtained from the sign of the functional equation:

$$\Lambda(E,s) = \pm \Lambda(E,2 - s)$$

Where $$\Lambda$$ is related to the $$L$$ function by: $$\Lambda(E,s) = N^{s/2} (2\pi)^{-s} \Gamma(s) L(E,s)$$ where $$N$$ is the conductor of $$E$$ over $$\mathbb{Q}$$.

(I'm still a little confused on the detailed definitions. Another reference, with slightly different definition relating $$\Lambda$$ and $$L$$, http://www.math.harvard.edu/~gross/preprints/ell2.pdf )

Anyway, the expansion definition of $$L$$ does not look like it would be valid for both sides of the functional equation, which would prevent just evaluating $$\Lambda$$ to check the sign. And due to the speed, I'm guessing Sage isn't evaluating $$L$$ at all here (or can immediately tell just by looking at a couple terms in the expansion?).

Is there some trick that allows extracting the sign without evaluating $$L$$?

• Maybe there is much easier but modularity theorem says $f(z)=\sum_na_ne^{2i\pi nz}$ is modular eigen form in $S_2(\Gamma_0(M))$ for every $Nd|M$. Changing finitely many $a_p$ you can make $M$ smaller (which means changing model of $E$ and taking isogeny). Then $\Lambda(E,s) = \pm \Lambda(E,2 - s)$ is true when $M=N$ the least possible that is $f$ is a newform. Once you know it as explained in homepages.warwick.ac.uk/~masgaj/book/fulltext/index.html you can compute the periods of the jacobian of $X_0(M)$ and action of Fricke's involution on it, $f,E$ will correspond to one of the periods – reuns Jan 29 at 21:49
• Note (assuming you are given a Neron model for $E$) Hasse bound says $|a_p| = p^{1/2}, a_n = O(n^{1/2+\epsilon})$ so $L(s,E) = \sum_{n \ge 1} a_n n^{-s}$ converges only for $\Re(s) > 3/2$ and you can't look directly at $\frac{\Lambda(s,E)}{\Lambda(2-s,E)}$ to find the sign of the functional equation. However (assuming you are given a Neron model of $E$ so that $f(z) = \sum_{n \ge 1} a_n e^{2i\pi nz}$. is a newform) you can look instead at $\frac{f(z)}{(Nz)^{-2}f(-1/(Nz))}$ – reuns Jan 29 at 22:10
• I don't know what Sage does, but 2 possible ways are to use the approximate functional equation or compute the global root number as a product of local root numbers. – Kimball Jan 30 at 18:24
• @Kimball how is a local root number calculated? – PluckyBird Jan 30 at 18:51
• @PluckyBird: Take a look at sections 1-3 of this paper. To summarize, the local root number at $\infty$ is always -1, +1 at all primes of good reduction, and -1/+1 at primes of multiplicative reduction depending on whether it's split or nonsplit, respectively. If your curve is semistable, then this is already enough. The case of additive reduction is a bit more complicated, but still reasonable to compute. – Brandon Carter Jan 31 at 5:23