Let $E$ be an elliptic curve and $E_n$ be its quadratic twist by $n$. Let $\phi_n: X_0(N_n) \to E_n$ be the normalized modular parametrization of $E_n$ ($\infty \to O$)

For some particular curves $E$, it seems (based on computing some examples in sage) to always be the case that $\phi_n(0) = O$ when the rank of $E_n$ is 0. For the curves I'm looking at, the torsion subgroup is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ for all of the $E_n$. What are some possible explanations for this or strategies to go about proving it?

Note: Edited to clarify the first comment.

  • $\begingroup$ I guess you normalize the modular parametrization by sending $\infty$ to the origin of $E$? So you're asking about the image of the winding element $[0]-[\infty] \in J_0(N_n)$ in $E_n(\mathbf{Q})$, is that right? I think this is a hard problem to determine the order of the winding element in full generality, but there might be partial results in special cases. $\endgroup$ – François Brunault Apr 3 '13 at 8:15
  • $\begingroup$ The order of $[0]-[\infty]$ in $J_0(N)(\mathbf{Q})$ has been determined by Ligozat, see Courbes modulaires de genre 1, Théorème 3.2.16 numdam.org/numdam-bin/fitem?id=MSMF_1975__43__5_0 As you will see the formula for general $N$ is quite complicated but maybe simplifies in your case $N=N_n$ for specific values of $n$. If this order happens to be odd then you're done. But in general you will have no choice but taking the arithmetic of $E_n$ into account, which amounts to understand how the winding element interacts with Hecke operators. $\endgroup$ – François Brunault Apr 3 '13 at 11:42
  • $\begingroup$ The formula for the BSD quotient as explained in g6hq's answer will give you an answer which is conditional on the truth of the full BSD formula. I'm not sure of the current state of the art but there might even already be results on the full BSD formula for quadratic twists with rank 0 of a given elliptic curve. $\endgroup$ – François Brunault Apr 3 '13 at 11:45

Note: this gives a plausible explanation, not a whole solution.

Hi, the BSD-quotient is $L(E_n,1)/\Omega_n = Sha\cdot\prod_p c_p(E_n)/|T|^2$. I think your question is mostly equivalent to asking why this quotient is an integer when twisting by $n$? The torsion is of size 4 (maybe there is an exceptional twist where it is larger, you did not specify). I think because of the full 2-torsion, the Tamagawa number should be 4 at any prime dividing $n$, except those dividing the original conductor. So if $n$ has at least two prime factors coprime to $N$ it seems you are done. When twisting by a prime or some $n$ with nontrivial gcd with $N$, then you need to take a more careful analysis, but the original curve itself should already have nontrivial Tamagawa numbers at some prime if indeed it has full 2-torsion. In short, the product of Tamagawa numbers outweighs the torsion in the denominator.

The definition of "0" here is just the cusp 0 on the boundary of the complex upper half plane. One is essentially integrating the modular form as $\int_0^{i\infty} f(z) dz$, to get the modular parametrization. But there are the constants like $2\pi$ and the Manin constant to worry about.

  • $\begingroup$ I calculated the Tamagawa numbers, and their product is always at least 16. There are a lot of primes of bad reduction for the curves I'm considering. Although I'm still interested in a proof that isn't conditional on BSD, this does give me an intuitive explanation. Thanks! $\endgroup$ – stl Apr 5 '13 at 1:44

I only know this material in a vague and impressionistic way, but I think that it the fact that $0$ goes to the identity on the elliptic curve follows from Eichler-Shimura theory.

If $E/\mathbb{Q}$ is parametrized by a modular curve, then it's a quotient of the Jacobian of the modular curve. In the composite map $$X_{0}(N) \to J_{0}(N) \to E$$ each arrow sends $0$ to $0$.

[Edit: Looking at this question I see that apparently 0 need not go to the identity in general. So now I'm puzzled. I think that my confusion might come from not knowing the definition of "0" on the modular curve in this context, which I was taking to be the preimage of the identity on the Jacobian.]


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