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!


1 Answer 1


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.

  • $\begingroup$ 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.) $\endgroup$ Nov 3, 2018 at 4:05
  • $\begingroup$ I've edited the answer to use a different example that meets your criteria. $\endgroup$ Nov 3, 2018 at 21:52
  • $\begingroup$ Thank you for this new example, Jeremy Rouse! You are right. $\endgroup$ Nov 4, 2018 at 21:06
  • $\begingroup$ 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. $\endgroup$
    – user262841
    Nov 5, 2018 at 9:49
  • $\begingroup$ @user262841 Thank you for the comment! Being holomorphic on $\mathbb{H}$ is significant. $\endgroup$ Nov 7, 2018 at 0:00

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