4
$\begingroup$

Let $E$ be an elliptic curve over $\mathbf{Q}$. Then we can base-change $E$ to $\mathbf{C}$ and apply the uniformization theorem to obtain: $$E(\mathbf{C}) \cong \mathbf{C}/(\mathbf{Z} + \mathbf{Z} \tau ) $$ for some complex number $\tau$ in the upper half plane. I've done a few numerical tests on Sage, and I've found that the real part of $\tau$ seems to be a rational number. So I wanted to ask: is it known to be true that if $E$ is defined over $\mathbf{Q}$ and $\tau$ is given as above, then $\rm{Re }\,\,\tau$ is a rational number? And if it is, would anyone be able to provide a reference for a proof?


Important note: an earlier version of the question said that $E$ was an elliptic curve defined over $\mathbf{C}$, not over $\mathbf{Q}$, which explains some of the comments. I've just edited it to say that $E$ must be defined over $\mathbf{Q}$.

$\endgroup$
5
  • 4
    $\begingroup$ Most certainly not always. $\tau$ can be literally any element of the upper half-plane, and not all of them have rational real parts. $\endgroup$
    – Wojowu
    Commented Oct 24, 2021 at 22:41
  • 1
    $\begingroup$ When you say you've done some numerical tests on Sage, what do you mean? $\endgroup$ Commented Oct 24, 2021 at 22:55
  • 3
    $\begingroup$ Ah, I should have mentioned that $E$ had to have been defined over the rationals! And then I'm considering the base change of $E$ to $\mathbf{C}$. Anyways, the answer to the question addresses this just fine. $\endgroup$ Commented Oct 25, 2021 at 0:06
  • 2
    $\begingroup$ Please do edit your question to mention the (very important!) feature that $E$ should have been defined over $\mathbb Q$ (and then base-changed to $\mathbb C$)... $\endgroup$ Commented Oct 25, 2021 at 0:18
  • $\begingroup$ Just edited the question. $\endgroup$ Commented Oct 25, 2021 at 0:27

1 Answer 1

8
$\begingroup$

However, if $E$ is defined over $\mathbb R$, then it's always possible to find a $\tau$ of the form either $\tau=ti$ or $\tau=\frac12+ti$ so that $E(\mathbb C)$ is analytically isomorphic (over $\mathbb R$, even) to $\mathbb C/(\mathbb Z+\mathbb Z\tau)$. So possibly the examples you were looking at are defined over $\mathbb R$, which you can check by seeing if $j(E)\in\mathbb R$.

Addendum If $E$ has complex multiplication, then $\mathbb Q(\tau)$ is an imaginary quadratic field. If $E$ does not have CM, then my recollection is that $\tau$ is transcendental over $\mathbb Q$. There is further information in the answer to When is the period of elliptic curve over the rationals transcendental?

$\endgroup$
3
  • 1
    $\begingroup$ I think it should be $\tau=ti$ or $\tau=\frac{1}{2}+ti$. $\endgroup$
    – Wojowu
    Commented Oct 24, 2021 at 23:17
  • $\begingroup$ Thank you! One follow-up: if $E$ is defined over the rationals, then can anything stronger be said about our choice of $\tau$? E.g- can we say anything about the imaginary part of $\tau$ as well? Or is this the best that we can say in general? $\endgroup$ Commented Oct 25, 2021 at 0:09
  • 1
    $\begingroup$ @chaad An elliptic curve over $\mathbb{R}$ can be approximated by elliptic curves over $\mathbb{Q}$, so one cannot say more in general. $\endgroup$ Commented Oct 26, 2021 at 7:48

You must log in to answer this question.

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