5
$\begingroup$

Let $E/\mathbb{Q}$ be an elliptic curve of rank 0, with modular parametrization $\phi: X_0(N) \to E$. Let $\Omega_0$ be the least positive real period. A paper I'm reading (Yoshida, Some variants of the congruent number problem I) seems to use the following line of reasoning. Say $\Omega_0/L(E/\mathbb{Q}, 1) = c$. We have that $\Omega_0/c = L(E/\mathbb{Q}, 1) = -I(0)$, so $I(0)$ must be a torsion point of exact order $c$.

I'm confused because from Cremona's tables, there are elliptic curves (for example 11a3) with torsion subgroup of order 5 but $c = 25$. So something is wrong here... Any idea what it is? I'm most likely just misinterpreting the paper or incorrectly generalizing (he was only doing this for a few specific elliptic curves).

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

Here is an expanded version of g6hq's answer. Your question is indeed sensitive to the Manin constant of the modular parametrization $X_0(N) \to E$. This in turns depends on whether the elliptic curve $E$ is the so-called "strong Weil curve" in its isogeny class, which by definition means that the kernel of $\phi_* : J_0(N) \to E$ is connected.

For example when $E=11a1=X_0(11)$, then $\phi$ is an isomorphism so everything is fine. But when $E=11a3=X_1(11)$, the modular parametrization $\phi : X_0(11) \to X_1(11)$ of least degree has degree 5 and in fact satisfies $\phi^* \omega_E = c_E \omega_f$ where $f=f_E \in S_2(\Gamma_0(11))$ is the newform associated to $E$, with a non-trivial Manin constant $c_E=5$. Thus the torsion point $I(0)$ has, in fact, order $c/c_E=5$.

In general, given any elliptic curve $E/\mathbf{Q}$, there is an optimal parametrization $\phi : X_1(N) \to E$. Such a parametrization conjecturally satisfies $\phi^* \omega_E = \omega_f$ (Stevens' conjecture). But the class of the divisor $[0]-[\infty]$ in the Jacobian of $X_1(N)$ is defined only over $\mathbf{Q}(\mu_N)^+$, not over $\mathbf{Q}$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I'm now trying this for a different curve, and am still confused. Let $E$ be the curve with label 70560bg1. It has $c_E = 1$, and $c = \Omega_0/L(E/\mathbb{Q}, 1) = 4$. But the torsion is just $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, so it can't have a torsion point of order $c/c_E = 4$. Is there something else that is messing it up in this case? $\endgroup$
    – stl
    Commented Mar 27, 2013 at 15:58
  • 1
    $\begingroup$ Just to be sure, are you considering $E=[0, 0, 0, -360297, 83126736]$? I'm finding $L(E,1)=16\Omega_0$ for this curve. Assuming the Manin constant is 1 (which seems to be the case according to Cremona's tables), this just means in terms of cycles that $\phi_* \{0,\infty\}=16 \gamma_E$ where $\gamma_E$ is a generator of $H_1(E,\mathbf{Z})^+$ (fixed subspace of homology under complex conjugation). This implies that $\phi(0)=\phi(\infty)=0$. In general, the denominator of $c$ (or $c/c_E$) will give you the exact order of $\phi(0)$. Here $c=16$ so the denominator is 1. $\endgroup$
    – François Brunault
    Commented Mar 28, 2013 at 9:22
  • $\begingroup$ Sorry, what I denoted by $c$ in the above comment is $1/c$ with your terminology... $\endgroup$
    – François Brunault
    Commented Mar 28, 2013 at 9:23
  • $\begingroup$ My confusion was stemming from thinking of $\Omega$ as always bigger than $L(E, 1)$, and therefore misreading the table for 70560bg1 as giving $\Omega/L(E, 1) = 8$ instead of $L(E, 1)/\Omega = 8$. Thanks for the help! $\endgroup$
    – stl
    Commented Apr 2, 2013 at 4:10
1
$\begingroup$

Hi, there is a nontrivial Manin constant of 5 for 11a3, I think this changes the answer, when comparing the L-value and periods.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .