Introduction

I. Moonshine functions: Dedekind eta quotients are very interesting, especially the class called moonshine functions (terminology after McKay, Ronan, Carnahan, et al). Conway and Norton found in 1979 that these functions span a linear space of \(172-9 = \color{red}{163}\) dimensions which McKay, in an email, remarked was a "delicious coincidence". There is growing evidence that each of these 163 dimensions contain a Ramanujan-type pi formula family, hence my interest in these functions.

II. Algebraic identities: First, some basics which will be useful later. Let \(a+m = b,\; a+n = c\) for some rational \(m,n\). Then the four trinomial identities of S1 are satisfied,
$$\begin{align}
a+m &= b\\
a+n &= c\\
a(m-n)+bn &=cm\\
b-c &= (m-n)\end{align}$$ These are responsible for quadruples of some trinomial identities found by Somos, as well as all for levels 6, 8, 10 (which only have four trinomial identities per level and prefixed by t6, t8, t10, respectively). The identities of S2 are also satisfied, $$\begin{align}
x_1 &= \frac{bc}{a} = a+\frac{mn}{a}+(m+n)\\
x_2 &= \frac{ac}{b} = b+\frac{m(m-n)}{b}+(-2m+n)\\
x_3 &= \frac{ab}{c} = c-\frac{n(m-n)}{c}+(m-2n)\end{align}$$This implies S3, $$x_0 = x_1 + \frac{(m-n)^2}{x_1} = x_2 + \frac{n^2}{x_2} + 2m= x_3 + \frac{m^2}{x_3} + 2n$$ and the linear relation S4, $$ (x_0 + r)+2a = x_1+x_2+x_3+r$$ where \(r\) may be chosen such that \(x_0 +r\) has nice properties, such as being a square. (Note that if \(m = -n\), then one can just set \(r = 0\) as \(x_0\) is already a square, a special case that will be relevant in certain levels.) These identities are valid for any \(a,b,c\) but become exceedingly interesting when all three \(a,b,c\) are eta quotients.

III. Example: Given the Dedekind eta function, $$\color{blue}{d_m = \eta(m\tau)}$$ where we suppress the argument \(\tau\) for ease of notation. Let,
$$a=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b = \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ then, $$\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ So \(m = -1,\,n=-9,\) and \(m-n = 8,\) and from \(S_2\) we get, $$\begin{align}
x_1 &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} = a+\frac{9}{a}-10\\
x_2 &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b}  = b-\frac{8}{b}-7\\
x_3 &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} = c+\frac{72}{c}+17\end{align}$$ where \(a,b,c\) can be expressed in terms of each other. To make \(x_2 + \frac{9^2}{x_2} -2\) a square, add \(r=20\) and we have the linear relation,
$$ \left(\sqrt{x_2} + \frac{9}{\sqrt{x_2}}\right)^2+2a = x_1+x_2+x_3+20$$ This is equivalent to one of the 9 linear dependencies in Conway and Norton’s 1979 paper, Monstrous Moonshine. For levels \(N = 6, 12, 10, 18, 30,\) those dependencies are, $$\begin{align}
j_{6A}+2\color{red}{j_{6E}} &= j_{6B}+j_{6C}+j_{6D}+20\\
j_{10A}+2\color{red}{j_{10E}} &= j_{10B}+j_{10C}+j_{10D}+8\\
j_{12A}+2\color{red}{j_{12I}}&= j_{12A}+j_{12B}+j_{12E}+8\\
j_{18B}+2\color{red}{j_{18D}} &= j_{18A}+j_{18C}+j_{18E}+4\\
j_{30B}+2\color{red}{j_{30G}}&= j_{30A}+j_{30C}+j_{30F}+5
\end{align}$$ Later we shall see this generalizes (but not completely) to other levels.

IV. Special property: The relevant triples {\(a,b,c\)} of level \(N = 6, 10, 12, 18, 30\) connected to the linear dependencies above are rather special since, using a common formula, they can generate another triple {\(a',b',c'\)} of level \(2N\) namely,
$$\begin{align}
a'(\tau) &= \sqrt{\frac{a(2\tau)\,c(\tau)}{b(\tau)}}\\
b'(\tau) &= \frac{b(2\tau)}{b(\tau)}\\
c'(\tau) &= \frac{a(\tau)\,c(2\tau)}{a'(\tau)\,b(\tau)}
\end{align}$$ where $$a+m = b,\quad a+n = c\\a'-m = b',\quad a'-n = c'$$So the highest it can generate is a level \(30\times2=60\) identity. However, for particular triples of level \(N = 10, 12, 18\), one can also generate a level \(4N\),
$$\begin{align}
a'(\tau) &= \sqrt{\frac{a(4\tau)}{a(2\tau)}}\\
b'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b^2(2\tau)\,c(\tau)}{a(4\tau)\,b(\tau)}}\\
c'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b(\tau)\,c^2(2\tau)}{a(4\tau)\,c(\tau)}}\end{align}$$ where $$a'\pm1 = b',\quad a'\mp1 = c'$$ and the \(\pm\) sign chosen appropriately. Thus, it can generate {\(a',b',c'\)} with level \(18\times4=72\) which is the highest I've observed. We will explicitly give these in the other posts.