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May 12, 2019 at 11:04 comment added François Brunault @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 $\Delta^{-1/12}$.
May 11, 2019 at 11:19 comment added MyNinthAccount It is probably better to bound $\Omega$ in terms of the discriminant than the conductor.
May 11, 2019 at 9:09 comment added Chris Wuthrich 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.
May 11, 2019 at 9:01 comment added Chris Wuthrich 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 $1$.
May 11, 2019 at 3:40 review First posts
May 11, 2019 at 3:49
May 11, 2019 at 3:37 history answered MyNinthAccount CC BY-SA 4.0