Excercise on j-invariant

Question

I have this problem.

Given $j:\cal{H}\rightarrow\mathbb{C}$ the j-invariant function defined on the upper-half complex plane as $j=1728\frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}$, where $g_2,g_3$ is the usual notation for Eisenstein's series.

The question is: given $t\in\cal{H}$ s.t. $j(t)=z\in\mathbb{C}$, find a $t'\in\cal{H}$ s.t. $j(t)=\bar{z}$.

I know that $j$ is a modular function of weight $0$ and I know a theorem for modular functions of weight k orders of poles and zeros:

$ord_\infty (f)+1/3ord_{\zeta_3}(f)+1/2ord_i(f)+\sum_{p\in F/\{\zeta_3,i\}}ord_p(f)=k/6$ where F is the fundamental domain of the action of $\mathbb{P}GL_2(\mathbb{Z})$ on $\cal{H}$.

Given this result I know that such a $t'$ exists: $f(\tau)=j(\tau)-\bar{z}$ is modular function of weight $0$ with a 1 order pole at infinity. Hence it must have at least one zero in the fundamental domain.

I've tried to do some algebric manipulation on $j$ to find $t'$, such as find evaulating $j(it)$ or $j(1/z)$ but it does not seem to work.

Here's a post on Inverting the modular $J$ function, which I think could answer your question, but maybe there's a nice answer for solving the specific equation $j(t') = \overline{z}$. — Viktor Vaughn, May 31 '24 at 17:53
Hint: the $j$-invariant has a $q$-expansion with integer coefficients, where $q = e^{2\pi i t}$. How do you turn $q$ into $\overline{q}$? — CJ Dowd, May 31 '24 at 18:00

score 1 · Answer 1 · answered Sep 12 '24 at 08:59

I will present two solutions. We claim that the answer is $-\bar{\tau}$. The main claim is that $\overline{j(\tau)}=j(-\bar{\tau})$. Then we have $j(-\bar{\tau})=\overline{j(\tau)}=\bar{z}$ since $j(\tau)=z$ and $-\bar{\tau}\in \mathbb{H}$ so the $j$ invariant makes sense.

We prove that for the Eisenstein series $E_{2k}(\tau)=\sum_{m,n}'(m+n\tau)^{-2k}$ (where ' indicates we don't sum $(0,0)$ and $k\in\mathbb{N}_{>1}$) satisfies $\overline{E_{2k}(\tau)}=E_{2k}(-\bar{\tau})$ which then our claim follows because $g_2$ and $g_3$ are scalings of $E_4, E_6$ and $j$ is a combination of $g_2$ and $g_3$. Taking conjugates in the definition of Eisenstein series we get $$ \overline{E_{2k}(\tau)}=\sum_{m,n}'\overline{(m+n\tau)^{-2k}}=\sum_{m,n}'(m+n\bar{\tau})^{-2k}=\sum_{m,l}'(m-l\bar{\tau})^{-2k}=E_{2k}(-\bar{\tau})$$ where we changed the variables $l=-n$. Thus the proof is completed.
We start with the Fourier expansion $j(\tau)=\sum_{n=-1}^{\infty} c_n e^{2\pi i n \tau}$ where $c_n$ are integers, hence real. Taking conjugates, we get $$\overline{j(\tau)}=\sum_{n=-1}^{\infty} \overline{c_n e^{2\pi i n \tau}}=\sum_{n=-1}^{\infty} c_n e^{-2\pi i n \bar{\tau}}=j(-\bar{\tau})$$ which concludes the proof.

Excercise on j-invariant

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