1
$\begingroup$

Let $E$ be the elliptic curve $$y^2 =x^3 - 19*67 x$$ and $P=[26011/625,2159616/15625]$, I want to compute $\hat{h}(P)$ using formula given in

Fujita, Y., & Terai, N. (2011). Generators for the elliptic curve $ y^ 2= x^ 3-nx$. Journal de théorie des nombres de Bordeaux, 23(2), 403-416.

To compute Archimedean part (formula (3.2) in the article), we should to compute $\log{\theta}$, there is a formula in the article: $$\theta=\sum_{k=0}^{\infty}{(-1)^kq^{k(k+1)/2}sin((2k+1)\frac{2\pi}{\omega_1}Re(z))}$$ Is there a command in Pari/GP or Magma to compute this value? As mentioned in the article, $|\theta|<1/(1-q)$, is there a lower bound for $|\theta|$ in elliptic curves of the form $y^2=x^3-nx$?

$\endgroup$
2
  • 1
    $\begingroup$ The formula for $\theta$ is (up to a scaling and a change of variables) a value of the Jacobi theta function, which is implemented in both PARI/GP and Magma (theta in GP, JacobiTheta in Magma). PARI/GP and Magma can also both compute the canonical height directly (ellheight in GP, CanonicalHeight in Magma). $\endgroup$ Nov 30, 2014 at 18:01
  • $\begingroup$ In Magma: E := EllipticCurve([-19*67,0]); pt := E![26011/625, 2159616/15625]; CanonicalHeight(pt); You can try this out with the online Magma calculator at magma.maths.usyd.edu.au/calc $\endgroup$ Nov 30, 2014 at 18:31

1 Answer 1

4
$\begingroup$

Is there some reason that you want to use this formula? You can get the height directly in PARI with the commands:

gp > EE=ellinit([0,0,0,-19*67,0]);

gp > PP=[26011/625,2159616/15625];

gp > ellheight(EE,PP)

%1 = 8.6541981

For the local archimedean height, the following formula is in Advanced Topics in the Arithmetic of Elliptic Curves Theorem VI.3.4. This uses the product form of the Weierstrass sigma function (theta function), rather than the series that you're using, but it still converges quite rapidly.

Theorem Let $u=e^{2\pi i z}$ and $q=e^{2\pi i\tau}$. Then $$ \lambda(z)=-\frac12B_2\left(\frac{\operatorname{Im} z}{\operatorname{Im}\tau}\right) \log|q|-\log|1-u|-\sum_{n\ge1}\log|(1-q^nu)(1-q^nu^{-1})|, $$ where $B_2(T)=T^2-T+\frac16$.

Regarding your question about a lower bound for your $\theta$, I think that the answer is no. In your notation, you have $\lambda(z)=-\log|\theta(z)|$, so your upper bound for $|\theta(z)|$ gives a lower bound for the height. But the height has a logarithmic pole at the identity element, and if $E(\mathbb{Q})$ has rank at least one, then the rational points are dense in the (identity component of the) real points. So in your case there are multiples of your point that are arbitrarily close to the identity (in the real topology), hence their archimedean heights can be arbitrarily large.

Using results from linear forms in logs, one can get some sort of upper bound for $\lambda(mz)$ that depends on $m$ and $z$ and $\tau$, but the limsup of this bound over $m$ is $\infty$.

$\endgroup$
3
  • $\begingroup$ Thanks for your answers, they are helpful. But I'm working on a problem and I need some quantities which appear in the formula (3.2) in the article, so I have to use this formula. I know Pari and Magma can compute the canonical height but I need to compute them directly. Unfortunately, when I use theta(exp(−2∗Pi),ellpointtoz(P)) the height which is obtained directly is different with the canonical height which is computed by Pari!!! – $\endgroup$ Dec 1, 2014 at 6:46
  • $\begingroup$ Is there a relation between $\theta(P)$ and $\theta(2P)$? $\endgroup$ Dec 1, 2014 at 11:46
  • 2
    $\begingroup$ @somayehdidari Up to possible small adjustments(and assuming that your $\theta$ is the function that I think it is), the function $\theta(2P)/\theta(P)^4$ is an elliptic function that vanishes at the 2-torsion points and has a triple pole at 0, so it is a multiple of $\wp'(z)$, the derivative of the Weierstrass $\wp$-function. $\endgroup$ Dec 1, 2014 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.