For an elliptic curve $E$ over $\mathbb{Q}$, it is not very difficult to show $L(E,1)\not=0$ (when the analytic rank$=0$) by computing the several Fourier coefficients but seem to be difficult to determine whether one has $L(E,1)=0$ (when the analytic rank$\not=0$). Is there a small constant $c$ such that, if we have $\mid L(E,1)\mid <c$, one can obtain $L(E,1)=0$ ?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Just to pick up on something mentioned in @MyNinthAccount's answer:
If you just want to determine whether or not $L(E, 1) = 0$, then there is another approach, which doesn't involve computing $L(E, 1)$ numerically to high precision and then arguing with inequalities. The point is that the ratio $$\frac{L(E, 1)}{\Omega^+_E}$$ is not only rational, but is exactly computable as a rational number (using modular symbols). So there is an algorithm which will determine, rigorously and in finitely many steps, whether the analytic rank is 0 or not. The standard number theory software packages (Sage, Magma, maybe also Pari) have built-in functionality to do this.
If you find by this method that $L(E, 1) = 0$, then determining what the analytic rank actually is is a much harder problem.
answered May 12, 2019 at 8:10
$\begingroup$
$\endgroup$
4
There is not a "constant" $c$ per se, but rather an expression in terms of the real period. This follows for instance from Manin's modular symbols theory, that the central $L$-value is a rational multiple (of bounded height) of the real period. More explicitly, you need to include the torsion and Tamagawa number contributions.
Using Birch and Swinnerton-Dyer as a guide, one has $${L(E,1)\over\Omega}{|T|^2\over \prod c_p}$$ is an integral square (possibly zero), and if it is less than 1 it indeed must vanish (one might need extra factors from Manin constants, I'm not sure).
So there is an implication: $$L(E,1)<{\prod_p c_p\over |T|^2}\Omega_{\rm re}\implies L(E,1)=0.$$
answered May 11, 2019 at 3:37
-
2$\begingroup$ That sounds like a conjectural bound, though recent work on BSD proves this almost. One can use the theorem of Manin and Drinfeld to get a better bound. Let $c_0$ be the Manin constant of the strong Weil curve in the isogeny class of $E$. Then $L(E,1)$ is in $\Omega/2/ \vert T\vert /c_0\, \mathbb{Z}$. For semistable curves $c_0$ is known to be $1$ or $2$ for instance, in general it is conjectured to be $1$. $\endgroup$ Commented May 11, 2019 at 9:01
-
3$\begingroup$ If you are after a more absolute bound $c$, then the question becomes what the smallest $\Omega$ are. By a conjecture of Goldfeld there is a $k$ such that $\Omega = O(N^{-k})$ as the conductor $N$ increases. The abc conjecture is equivalent to $k=1/2+\varepsilon$ if I remember correctly. Maybe $\Omega> 1/N$ for all curves, I would not know. $\endgroup$ Commented May 11, 2019 at 9:09
-
$\begingroup$ It is probably better to bound $\Omega$ in terms of the discriminant than the conductor. $\endgroup$ Commented May 11, 2019 at 11:19
-
1$\begingroup$ @ChrisWuthrich There are convexity bounds of the form $L(E,1) \ll N^{1/4}$ if I recall correctly, so together with the bound you mention (and assuming $c_0=1$), this seems to prove $\Omega \ll N^{1/4}$. Maybe it is possible to get a lower bound for the real period using its AGM expression as in Watkins, Some heuristics on elliptic curves, section 3.3. Watkins mentions that $\Omega$ is very roughly $\Delta^{-1/12}$. $\endgroup$ Commented May 12, 2019 at 11:04