The j-function of a point on the boundary of the fundamental domain is real valued.

Question

I am trying to prove the following statement:

Let $z \in \mathbb{D} $ (the standard fundamental domain for $SL_{2}(\mathbb{Z})$). Prove that if $z$ lies on the boundary of $\mathbb{D}$, or if $Re(z)=0$, then $j(z) \in \mathbb{R}$.

So far I have shown that $j(i)=1728$ and $j(e^{(2\pi i)/3})=0$ but I am not sure how to proceed...

Answer 1

You can use the moduli interpretation of the $j$-invariant to figure out why $j$ should be real there. If you think of $j(\tau)$ as $j(\mathbb C/\mathbb Z + \tau \mathbb Z)$, you see that $j(\tau)$ is real precisely when the elliptic curve $\mathbb C/\mathbb Z + \tau \mathbb Z$ admits a model over the real numbers. If $\mathbb C/\mathbb Z + \tau \mathbb Z$ admits a model over $\mathbb R$, then it is isomorphic to its complex conjugate $\mathbb C/\mathbb Z + \overline{\tau} \mathbb Z$, which is equivalent to the requirement that the lattice $\mathbb Z + \tau \mathbb Z$ is homothetic to its complex conjugate. So, to summarize, $j(\tau) \in \mathbb R$ precisely when there exists $\alpha \in \mathbb C^\times$ such that

$$\mathbb Z + \tau \mathbb Z = \alpha(\mathbb Z + \overline{\tau} \mathbb Z).$$

This happens for instance when $\Re \tau$ is a half-integer, because if $\tau = a/2 + yi$,

$$\overline{\tau} =\overline{a/2 + yi} = a/2 - yi = a - \tau \in \mathbb Z + \tau \mathbb Z$$

so the two lattices are even equal - we can take $\alpha = 1$. Thus $j$ is real on the vertical half-lines $\Re \tau = a/2$. It also happens along $|\tau| = 1$; because then $\tau\overline{\tau} = 1$, and we have

$$\mathbb Z + \tau \mathbb Z = \tau(\mathbb Z + \overline{\tau} \mathbb Z).$$

So $j$ is real along the upper half of the unit circle.

Answer 2

Hint: Show that if $f(\tau)=g(\exp(\alpha\mathrm{i}\tau))$ for some Laurent series $g$ with real coefficients and real $\alpha>0$, then $f(-\bar{\tau}) = \bar{f}(\tau)$. Then use the symmetries of $j$ to show that on the boundary of the fundamental domain, as well as for $\operatorname{Re}\tau=0$, you get $j(\tau) = j(-\bar{\tau}) = \bar{j}(\tau)$.