Define the following,

$$j(\tau) = \Big(\tfrac{E_4(\tau)}{\eta^8(\tau)}\Big)^3 = {1 \over q} + 744 + \color{blue}{196884} q + 21493760 q^2 + 864299970 q^3 + \cdots \tag{1}$$

$$j_{2A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 = \tfrac{1}{q} + 104 + \color{blue}{4372}q + 96256q^2 + 1240002q^3+\cdots \tag{2}$$

$$j_{3A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(3\tau)}\big)^{6}+3^3 \big(\tfrac{\eta(3\tau)}{\eta(\tau)}\big)^{6}\Big)^2 = \tfrac{1}{q} + 42 + \color{blue}{783}q + 8672q^2 +65367q^3+\dots \tag{3}$$

$$j_{4A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(4\tau)}\big)^{4}+4^2 \big(\tfrac{\eta(4\tau)}{\eta(\tau)}\big)^{4}\Big)^2 = \tfrac{1}{q} + 24+ \color{blue}{276}q + 2048q^2 +11202q^3+\dots\\ \tag{4}$$ $$j_{7A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(7\tau)}\big)^{2}+7 \big(\tfrac{\eta(7\tau)}{\eta(\tau)}\big)^{2}\Big)^2 = \tfrac{1}{q} + 10+ \color{blue}{51}q + 204q^2 +681q^3+\dots\\ \tag{5}$$

where $\eta(\tau)$ is the Dedekind eta function.

(Note that $196883, 4371, 782, 276, 51$ are degrees of the irreducible representations of the Monster, Baby Monster, Fischer Fi23, Conway Co1, and Held groups, respectively.)

The asymptotic formula (by Rademacher?) for the coefficient of $q^n$ of $(1)$ is given by,

$$a(n) \approx \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\,n^{3/4}}\tag{for 1}$$

For example, let $a(n)$ be the coefficient, and $a'(n)$ given by the formula, then,

$$\begin{array}{cccc} n&100&200&300\\ a(n)&8.38\,\text{x}\,10^{52}&2.011\,\text{x}\,10^{75}&3.293\,\text{x}\,10^{92}\\ a'(n)&8.40\,\text{x}\,10^{52}&2.016\,\text{x}\,10^{75}&3.299\,\text{x}\,10^{92}\\ \end{array}$$

Question: What are the analogous coefficient formulas for $(2), (3), (4), (5)$?

$\color{blue}{\text{Edit}}$: After perusing the OEIS, it seems that the fourth has,

$$d(n) \approx \frac{e^{2\pi\sqrt{n}}}{2\,n^{3/4}}\tag{for 4}$$

$$\begin{array}{cccc} n&100&200&300\\ d(n)&3.04\,\text{x}\,10^{25}&3.64\,\text{x}\,10^{36}&1.26\,\text{x}\,10^{45}\\ d'(n)&3.06\,\text{x}\,10^{25}&3.66\,\text{x}\,10^{36}&1.27\,\text{x}\,10^{45}\\ \end{array}$$

though I have no proof that this is its correct asymptotic formula.


In a recent paper with Ken Ono and John Duncan, we give exact formulas for the coefficients of the Monsterous Moonshine modules. I will give a link to a preprint a few days when I can, but the relevant part relies on prior work of Bringmann and Ono (http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/115.pdf) which gives exact formulas for coefficients of modular forms assuming knowledge of their poles at cusps.

Each of your $j_g(\tau)=\sum a_g(n)q^n$ is a Hauptmodul for a group of the form $\Gamma_g=\left\langle \Gamma_0(N),\begin{pmatrix}0&-1\\N&0\end{pmatrix}\right\rangle,$ where $N=1,2,3,$ or $4$. In this case, the dominant term in the exact formulas for $a_g(n)$ is $$\frac{2\pi}{\sqrt{N n}}I_1\left(4\pi \sqrt{n/N}\right)\sim \frac{e^{4\pi \sqrt{n/N}}}{\sqrt{2}N^{1/4}n^{3/4}}.$$ Here $I_k$ is the usual $I$-Bessel function.

  • $\begingroup$ Ah, excellent. A simple general form! (P.S. I added $N=7$ just for good measure. $N=5,6$ have slightly different eta quotients.) $\endgroup$ – Tito Piezas III Nov 19 '14 at 2:43
  • $\begingroup$ The paper I referenced can be accessed here: arxiv.org/abs/1411.6571 $\endgroup$ – Michael Griffin Dec 2 '14 at 15:38
  • $\begingroup$ Thanks. By the way, do you know the reason behind the Conway-Norton observation why the moonshine functions span a linear space of dimension 163? McKay was kind enough to refer to me a few years ago Gukov, Vafa's paper (arxiv.org/abs/hep-th/0203213) where in p.21 they discuss something similar. $\endgroup$ – Tito Piezas III Dec 2 '14 at 16:06

If we write $\psi(\tau) = \frac{\Delta(\tau)}{\Delta(2\tau)} = \sum_{n=-1}^\infty s(n) q^n$ and $\phi(\tau) = \frac{\Delta(2\tau)}{\Delta(\tau)} = \sum_{n=1}^\infty b(n) q^n$, then $j_{2A}(\tau) = \psi(\tau) + 128 + 2^{12}\phi(\tau)$. Asymptotics for $|s(n)|$ and $b(n)$ and explicit upper bounds are given in the paper http://arxiv.org/pdf/1408.1083.pdf, written with Kyle Pratt.

Specifically, we have $$|s(n)| \sim \frac{e^{2\pi\sqrt{n}}}{2n^{3/4}},$$ $$b(n) \sim \frac{2^{1/4}}{8192}\frac{e^{2\pi\sqrt{2n}}}{n^{3/4}}.$$

  • $\begingroup$ Ah, so I was right about the asymptotics of $j_{4A}(\tau)$. We have $$j_{4A}(\tau)= \tfrac{1}{q} + 24+276q + 2048q^2 +11202q^3 + 49152q^4 +\dots$$ $$\begin{aligned}\psi(\tau) &= \left(\tfrac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\\ &= \tfrac{1}{q} - 24+276q - 2048q^2 +11202q^3 - 49152q^4 +\dots\end{aligned}$$ and their coefficients grow as $$|d(n)|=|s(n)| \sim \frac{e^{2\pi\sqrt{n}}}{2n^{3/4}}$$ $\endgroup$ – Tito Piezas III Nov 13 '14 at 2:48

The coefficients of a modular form of non-positive weight can be given by an explicit formula that depends only on the poles at cusps (and constant terms when the weight is zero). The asymptotics are given by asymptotics of modified Bessel functions, which are well-known. Rademacher gave the formula in the level 1 case, and higher level cases are subsumed by inducing to a level 1 vector-valued form.

You can find an explicit formulation as Theorem 1.9 in Giménez's dissertation. While the formula in full generality is rather complicated, many of the terms will vanish for the case at hand, since you are working with Hauptmoduln of low level.

You can also extract the answer from earlier, slightly less explicit descriptions in section 2 of Dijkgraaf-Maldacena-Moore-Vafa, A Black Hole Farey Tail and Lemma 5.3 of Borcherds, Automorphic forms on $O_{s+2,2}(\mathbf{R})$ and infinite products

  • $\begingroup$ Thanks for the detailed references. As you point out, the general formula is rather complicated. Hopefully, someone will be able to translate it for the remaining case $j_{3A}(\tau)$. $\endgroup$ – Tito Piezas III Nov 13 '14 at 2:57

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