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For any $n\geq 1$, let $E_n $ be the elliptic curve given by the equation $y^2 = x(x-1)(x-\zeta_{15^n})$. Here $\zeta_{m} = \exp(2\pi i /m)$ for any positive integer $m$.

There is a unique element $\tau_n$ in the complex upper half plane such that $E_n = \mathbf{C}/(\mathbf{Z} + \tau_n\mathbf{Z})$.

I need that $\mathrm{Im}(\tau_n) > \frac{1}{2}$. Can we show this?

It might not be true, but if it is it would help me a lot.

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$\tau_n$ is only unique up to the action of $SL(2,\mathbb{Z})$. If you choose it from the standard fundamental domain then $Im \tau_n\geq\sqrt{3}/2>1/2$. –  user8268 Oct 13 '11 at 21:54
    
If you have an ell. curve in Legendre form $y^2 = x(x-1)(x-\lambda)$ and $\lambda$ is close to $0$ (or $1$), why can we choose $\tau$ such that its imaginary part is greater than 1/2? –  shaye Oct 14 '11 at 7:29

1 Answer 1

More or less by definition, $\tau(\lambda)$ is the ratio of the periods of $E(\lambda):y^2=x(x-1)(x-\lambda)$. The periods are the integrals of $dx/y$ over a pair of generators for $H_1(E_n, \mathbf{Z})$. Ambiguity in the choice of generators for the homology of $E$ implies that $\tau(\lambda)$ is only well-defined up to the action of an element of $\text{SL}_2(\mathbf{Z})$.

The periods are given, for example, by

$$\left\{2\int_0^1 \frac{\mathrm d x}{y}, 2\int_1^\lambda \frac{\mathrm dx}{y}\right\}.$$

These can be expressed in terms of the hypergeometric function using Euler's integral representation of the hypergeometric function.

Finding special values of $\tau$, I suppose, is then equivalent to finding special values of the hypergeometric function. You can use a CAS to get an arbitrarily good approximation, but I'm in no position to give an exact value of your $\tau(E_n)$ as a function of $n$.

Where did you get this problem from?

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Or even better, one can use the arithmetic-geometric mean (or variations of it) to express those two periods. –  J. M. Dec 4 '11 at 5:20

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