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.