Entry 44

Level 12 (Part 1). We found one triple $(a,b,c)$ each for Levels 6, 8, 10. But for Levels 12, 18, 24, we can find more than one triple, which complicates things. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2+1=\left(\frac{d_3^3\,d_4}{d_1\,d_{12}^3}\right)\\ \quad\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2-3=\left(\frac{d_1^3\,d_4\, d_6^2}{d_2^2\,d_3\,d_{12}^3}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a-3=c$$ so $(m,n)=(1,-3)$. Then  $$\frac{ab}{c} = \left(\frac{d_3\, d_4}{d_1\,  d_{12}}\right)^{4}\\ \frac{bc}{a} = \left(\frac{d_1\, d_3}{d_4\,  d_{12}}\right)^{2}\\ \quad\frac{ac}{b} =  \left(\frac{d_1\, d_4\,d_6}{d_2\,d_3\,  d_{12}}\right)^4\\ \qquad d = \left(\frac{d_2^2\, d_6^2}{d_1\,d_3\,d_4\,d_{12}}\right)^6$$ where $(a,b,c)$ are the McKay-Thompson series of class 12I (A187144, A187130) and $d$ is the McKay-Thompson series of class 12A (A112147). They obey $$\left(\sqrt{\frac{16a}{bc}}+\sqrt{\frac{bc}{a}}\right)^2=d$$ $$d+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}+8$$ Equivalently, $$\left(\frac{16a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ which is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$. However, for levels $(12, 18, 24)$ we can use a fourth eta quotient $d$ to simplify the linear dependency.