Entry 52

Level 24, Part 3. To summarize the two parts of level 24, we found two eta triples \((a,b,c)\) and \((d,e,f)\). Let \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\). Define $$a=\left(\frac{d_6\, d_{12} }{d_2\, d_ 4}\right) \left(\frac{d_1\, d_8 }{d_3\, d_{24} }\right)^2,\quad b=\left(\frac{d_1\, d_8 }{d_2\, d_4 }\right) \left(\frac{d_6\, d_ {12}}{d_3\, d_{24} }\right)^3,\quad c = \frac{\big( d_2\, d_4\big)^2 }{d_1\, d_3\, d_8\, d_{24} }\quad \\ d=\left(\frac{d_2\, d_{12} }{d_4\, d_ 6}\right) \left(\frac{d_3\, d_8 }{d_1\, d_{24} }\right)^2,\quad e=\frac{d_2^5\, d_3\, d_8\, d_{12}^5 }{\big(d_1\, d_4\, d_6\, d_{24}\big)^3},\qquad f=\frac{\big(d_4\, d_6\big)^4 }{d_1\, d_2^2\, d_3\, d_8\, d_{12}^2\, d_{24}}$$

then they obey a lot of relationships, $$\frac{a}{b c}= \frac{d}{e f}$$

$$a+1 = b,\quad a+3 = c\\ d+1=e,\quad d-1=f$$

$$\frac{ab}c+ \frac{bc}a+ \frac{ac}{b} - \left(\frac{de}f+ \frac{ef}d+ \frac{df}{e}\right) = 2(a-d) = 2(b-e) = 2(c-f-4)$$ as well as

$$\left(\frac{4a}{bc}+\frac{bc}{a}\right)+2a  = \color{red}{\frac{ab}c}+ \frac{bc}a+ \frac{ac}{b}\\ \left(\frac{4d}{ef}+\frac{ef}{d}\right)+2d  = \color{red}{\frac{de}f}+ \frac{ef}d+ \frac{df}{e}\\ \;\quad\left(\frac{4a}{bc}+\frac{bc}{a}\right)+\frac{2cd}{e} - 2 = \color{red}{\frac{ab}c}+\color{red}{\frac{de}f}+\left(\frac{cd}{e}-\frac{3e}{cd}\right)$$ and so on. Combining the last three relations by getting rid of the red non-moonshine terms yields the most complicated of the \(9\) dependencies found by Conway, Norton, and Atkin such that the monster functions span a linear space of \(172-9=163\) dimensions.