# Is $j(\tau)^{1/3}$ the hauptmodul for the congruence subgroup generated by $\tau\rightarrow\tau+3, \tau\rightarrow-1/\tau$?

The 3rd root of the modular invariant $$j$$ is $$j(\tau)^{1/3}=q^{-1/3}(1+ 248q+ 4124q^2+ 34752q^3+\cdots),$$ where $$q=e^{2\pi i \tau}$$.

I was wondering if $$j(\tau)^{1/3}$$ the hauptmodul for the congruence subgroup generated by $$\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$$.

If this true, can we say the following assertion? If a function $$f(\tau)$$ that takes the form $$f(\tau)=q^{-1/3}(1+\sum_{n=1}^{\infty} a_n q^n)$$ with $$a_n \geq 0$$ and is invariant under $$\tau \rightarrow \tau+3, \tau \rightarrow-1/\tau$$, then $$f(\tau)=j(\tau)^{1/3}$$.

Thanks a lot!

The function $$x(\tau) = j(\tau)^{1/3}$$ is a hauptmodul, just not for the group that you indicate. This function is also invariant under $$\tau \mapsto \frac{2 \tau + 1}{\tau + 1}$$ and $$\tau \mapsto \frac{\tau+1}{\tau + 2}$$, and this means that $$x(\tau)$$ is a hauptmodul for an index $$3$$ subgroup of $${\rm SL}_{2}(\mathbb{Z})$$, which is the normalizer of a non-split Cartan modulo 3. (This has been known for quite a while. If you want to read a proof of this, see the paper of Imin Chen titled "On Siegel's modular curve of level 5 and the Class Number One Problem" published in the Journal of Number Theory in 1999. Look in Section 4.) The subgroup you specify has index at least $$18$$ (and it may not even be congruence).
I think that the natural modification of your second question also has a negative answer. (Edited to fix the example.) In particular, $$x(\tau) \frac{j(\tau)}{j(\tau)-1728} = q^{-1/3} (1 + 1976q + 2133020q^{2} + \cdots)$$ is a modular function for the same group as $$x(\tau)$$ and also has non-negative Fourier coefficients. With some thought, one can see that $$j(\tau)$$ has non-negative Fourier coefficients, and so does $$\frac{1}{j(\tau) - 1728} = \frac{E_{4}^{3}/E_{6}^{2} - 1}{1728}$$. Here $$E_{4} = 1 + 240 \sum_{n=1}^{\infty} \sigma_{3}(n) q^{n}$$ and $$E_{6} = 1 - 504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$$ are the usual Eisenstein series.
• Thank you for the answer, Jeremy! Maybe I made a confusion in the second question. The powers of $f(\tau)=q^{-1/3}(1+\sum_{n=1}^{\infty} a_n q^n)$ are integer spaced, while $$\frac{x(\tau)^{2}}{x(\tau) - 12} = q^{-1/3}(1 + 12q^{1/3} + 144q^{2/3} + 1976q^{1} + \cdots)$$ is not. (I am afraid the q-expansion by you is not correct.) Nov 3, 2018 at 4:05
• I just add a comment regardless of Rouse's perfect answer. The conditions of Wu's second question determine the cube root of $j$ in some sense: among functions which are "holomorphic on $\mathbb{H}$", the cube root of $j$ is the unique function satisfying the conditions of the second question. Nov 5, 2018 at 9:49
• @user262841 Thank you for the comment! Being holomorphic on $\mathbb{H}$ is significant. Nov 7, 2018 at 0:00