# Can the pre-image of the real points of an elliptic curve under the modular parametrization be identified as points in the complex upper-half plane?

Consider an elliptic curve $E/\Bbb Q$ and let $\Phi:\Gamma_0(N)\backslash\overline{\mathfrak{H}}\rightarrow E(\Bbb C)$ be the analytical description of its modular parametrization. We know that this is a map defined over $\Bbb Q$ between algebraic curves over $\Bbb Q$, but in general for $\tau\in{\mathfrak{H}}$ the field of definition of the corresponding point in $X_0(\Bbb C)$ is not easy to identify.

Is it possible to identify points in $\mathfrak H$ that correspond to real points of $X_0(N)$ or points that map to real points of $E$?

Can anything be said about points mapping to real points under the Weierstrass parametrization $(\wp,\wp'):\Bbb C/\Lambda\rightarrow E(\Bbb C)$?

Thanks

I don't understand your first question, since the upper half plane has no real points. Or do you mean the points in $\mathfrak h$ that map to $E(\mathbb R)$?

For your second question, this is quite easy to answer. The easiest way to study it is to use the map $z\mapsto q=e^{2\pi iz}$ to identify $\mathbb C/\Lambda$ with $\mathbb C^*/q^{\mathbb Z}$, where $q=e^{2\pi i\tau}$. The assumption that $E$ is defined over $\mathbb R$ will give $q\in\mathbb R^*$, and it's easy to identify the inverse image of $E(\mathbb R)$ for the map $\mathbb C^*/q^{\mathbb Z}\to E(\mathbb C)$. For details, see for example Advanced Topics in the Arithmetic of Elliptic Curves, Section V.2. (There are some small subtleties depending on whether you want to classify elliptic curves defined over $\mathbb R$ up to $\mathbb C$-isomorphism, or up to $\mathbb R$-isomorphism.)

• Thanks a lot! Indeed the formulation was very unclear, I meant points of $\mathfrak{H}$ corresponding to real points of $X_0(N)$ or mapping to real points of $E$. I edited the question to make it clearer. Dec 21, 2014 at 9:55

Yes. If we take the invariant differential form of our elliptic curve and look at just the imaginary part, we get a real 1-form. The curves on which this 1-form vanishes are curves on which the invariant differential is real. One of these is the real locus.

So on the upper half plane, if we take the modular form, remember that it's a differential form, take the imaginary part, and look at its level sets, one of them will map to the real points of the elliptic curve.

Which level set? Well, it has to contain the rational points, and a cusp is a rational point of the modular curve, so it maps to a rational point of the elliptic curve. So you can start at a cusp.

The only tricky thing is ensuring that you normalize your modular form so it's defined over the reals as an algebraic differential form. I believe the usual normalization, with $q$ the leading term, accomplishes this.

I would be interested to see a picture of what this looks like for the modular curve of level 11, say.

• I guess you answer what the preimage of $X_0(N)(\mathbb{R})$ in the upper half plane looks like. Surely the imaginary axis belongs to it as complex conjugation on the curve comes from the reflection through the imaginary axis $\tau\mapsto -\bar\tau$. Dec 21, 2014 at 9:30
• Thanks Will! Do I understand correctly that this is what you're saying: you view $E$ as a quotient of the moduli space $X_0(N)$. Then the real locus maps to a real curve on $E(\Bbb C)$ with the property that the imaginary part of the invariant differential vanishes. On $\mathfrak H$ this differential is $2\pi if(\tau)d\tau$ up to a constant factor and the imaginary part of the differential vanishes on curves over which the imaginary part of the modular form is constant. Dec 21, 2014 at 10:46
• @ChrisWuthrich Is it easy to see that complex conjugation corresponds to reflection in the imaginary axis? Dec 21, 2014 at 10:51
• Coordinates on the affine model over $\mathbb{Q}$ are $j(\tau)$ and $j(N\tau)$. These are power-series in $q=e^{2\pi i \tau}$ with real coefficients. Now $\bar{q} = e^{2\pi i(-\bar\tau)}$ confirms that complex conjugation is the reflection on the imaginary axis. Dec 21, 2014 at 19:40
• @ChrisWuthrich Ah, of course, thanks Dec 21, 2014 at 23:10