These are not monster functions (with the exception of the first). But I'm including them since this seems to be the highest level such that the requirement of the triple relation works, namely that \(a,\,a+m,\,a+n\) are all eta quotients. For convenience, first define the eta quotient, \(\color{blue}{f_p} = \frac{d_p}{d_{2p}} =\frac{\eta(p\tau)}{\eta(2p\tau)}\). Let,
\begin{align}
a(\tau) &= \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right)\\
b(\tau) &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)f_6^2 \\
c(\tau) &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)f_6^2
\end{align} such that \(a+1 = b,\,a-1 = c\). Then define, $$\begin{align}
j_{12F}(\tau) &= f_6^4= \left(\frac{d_6}{d_{12}}\right)^4 \;=\; \frac{bc}a\\
k_{72A}(\tau) &= f_6^4+\frac{4}{f_6^4}\\
k_{72B}(\tau) &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)^2 = \frac{ac}b\\
k_{72C}(\tau) &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)^2 \;=\; \frac{ac}b\\
\color{red}{k_{72D}}(\tau) &= \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right)\; = \; a
\end{align}$$ Then we have an analogous linear relation,
$$k_{72A}+2\color{red}{k_{72D}} =k_{72B}+k_{72C}+ j_{12F}+2$$ Curiously, notice that the form of the last three \(k_{72}\) appear in other functions,
$$\begin{align}
\frac1{j_{36A}(\tau)} &= \left(\frac{d_1\, d_4\,d_6^2\, d_9\, d_{36}}{ d_2\,d_3^2\,d_{12}^2\, d_{18}}\right)^2\\
\frac1{k_{36A}(\tau)} &= \left(\frac{d_{3}^2\,d_{4}\, d_{36}}{d_{1}\, d_{9}\,d_{12}^2}\right)^2\\
\frac1{\sqrt{j_{18A}(2\tau)}} &=\left(\frac{d_4^2\,d_{6}^2\, d_{36}^2}{d_2\, d_{12}^4\,d_{18}}\right)\qquad\qquad
\end{align}$$
Thursday, October 31, 2019
Level 60
$$\begin{align}j_{60A}(\tau) & = j_{60E}(\tau)+\frac{1}{ j_{60E}(\tau)} = \sqrt{j_{30B}(2\tau)} \\
j_{60B}(\tau) &= \frac{(d_2\, d_{6}\, d_{10}\, d_{30})^2}{d_1\, d_{3}\, d_{4}\, d_{5}\,d_{12}\, d_{15}\, d_{20}\, d_{60}}\\
j_{60C}(\tau) &= \frac{a}{b} = \frac1{a}(d_6\,d_{10})^3+1 = \frac{a}{\,b^2}(d_2\,d_{30})^3-1\\
j_{60D}(\tau) &= \left(\frac{d_1\, d_{12}\, d_{15}\, d_{20}}{d_3\, d_{4}\, d_{5}\, d_{60}}\right)\\
&=\left(\frac{d_2\, d_6\, d_{10}\, d_{30}}{d_3\, d_{4}\, d_{5}\, d_{60}}\right)-1\\
j_{60E}(\tau) &= \frac{d_4\, d_6\, d_{20}\, d_{30} } {d_2\, d_{10}\, d_{12}\, d_{60}} =\sqrt{j_{30D}(2\tau)}\\
j_{60F}(\tau) &= \left(\frac{d_{12}\, d_{30}}{d_6\, d_{60}}\right)
\end{align}$$ where
$$\begin{align}
a &= d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}\\
b &= d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}
\end{align}$$
There are no linear relations between these functions.
j_{60B}(\tau) &= \frac{(d_2\, d_{6}\, d_{10}\, d_{30})^2}{d_1\, d_{3}\, d_{4}\, d_{5}\,d_{12}\, d_{15}\, d_{20}\, d_{60}}\\
j_{60C}(\tau) &= \frac{a}{b} = \frac1{a}(d_6\,d_{10})^3+1 = \frac{a}{\,b^2}(d_2\,d_{30})^3-1\\
j_{60D}(\tau) &= \left(\frac{d_1\, d_{12}\, d_{15}\, d_{20}}{d_3\, d_{4}\, d_{5}\, d_{60}}\right)\\
&=\left(\frac{d_2\, d_6\, d_{10}\, d_{30}}{d_3\, d_{4}\, d_{5}\, d_{60}}\right)-1\\
j_{60E}(\tau) &= \frac{d_4\, d_6\, d_{20}\, d_{30} } {d_2\, d_{10}\, d_{12}\, d_{60}} =\sqrt{j_{30D}(2\tau)}\\
j_{60F}(\tau) &= \left(\frac{d_{12}\, d_{30}}{d_6\, d_{60}}\right)
\end{align}$$ where
$$\begin{align}
a &= d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}\\
b &= d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}
\end{align}$$
There are no linear relations between these functions.
Level 40
$$\begin{align}
j_{40A}(\tau) &= \sqrt{j_{20B}(\tau)} = \sqrt[4]{j_{10A}(2\tau)}\\
j_{40B}(\tau) &= \sqrt{j_{20A}(2\tau)}= \left(\frac{d_4^2\, d_{20}^2}{d_2\, d_8\, d_{10}\, d_{40}}\right)^2\\
&= j_{40C}(\tau)+\frac1{j_{40C}(\tau)}\\
j_{40C}(\tau) &= \sqrt{j_{20F}(2\tau)}= \left(\frac{d_8\, d_{10}}{d_2\, d_{40}}\right)
\end{align}$$
There is no linear relation between these functions. \(j_{40A}\) is one of the few that involves a 4th root.
j_{40A}(\tau) &= \sqrt{j_{20B}(\tau)} = \sqrt[4]{j_{10A}(2\tau)}\\
j_{40B}(\tau) &= \sqrt{j_{20A}(2\tau)}= \left(\frac{d_4^2\, d_{20}^2}{d_2\, d_8\, d_{10}\, d_{40}}\right)^2\\
&= j_{40C}(\tau)+\frac1{j_{40C}(\tau)}\\
j_{40C}(\tau) &= \sqrt{j_{20F}(2\tau)}= \left(\frac{d_8\, d_{10}}{d_2\, d_{40}}\right)
\end{align}$$
There is no linear relation between these functions. \(j_{40A}\) is one of the few that involves a 4th root.
Level 36, part 1
$$\begin{align}
j_{36A}(\tau) &= \left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\, d_4\,d_6^2\, d_9\, d_{36}}\right)^2\\
&=j_{36B}(\tau)+\frac{3}{j_{36B}(\tau)}+3\\
j_{36B}(\tau) &=\left(\frac{d_1\, d_4\, d_{18}}{d_2\, d_9\, d_{36}}\right)\\
j_{36C}(\tau) &= \left(\frac{d_6^2\, d_{12}^2}{d_2\, d_4\,d_{18}\, d_{36}}\right) = \sqrt{j_{18B}(2\tau)}\\
j_{36D}(\tau) &= \left(\frac{d_4\, d_{9}}{d_1\, d_{36}}\right)
\end{align}$$
There is no linear relation between these functions.
j_{36A}(\tau) &= \left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\, d_4\,d_6^2\, d_9\, d_{36}}\right)^2\\
&=j_{36B}(\tau)+\frac{3}{j_{36B}(\tau)}+3\\
j_{36B}(\tau) &=\left(\frac{d_1\, d_4\, d_{18}}{d_2\, d_9\, d_{36}}\right)\\
j_{36C}(\tau) &= \left(\frac{d_6^2\, d_{12}^2}{d_2\, d_4\,d_{18}\, d_{36}}\right) = \sqrt{j_{18B}(2\tau)}\\
j_{36D}(\tau) &= \left(\frac{d_4\, d_{9}}{d_1\, d_{36}}\right)
\end{align}$$
There is no linear relation between these functions.
Level 30
$$\begin{align}
j_{30A}(\tau) &= \left(\frac{d_1\, d_6\, d_{10}\, d_{15}}{d_2\, d_{3}\, d_{5}\, d_{30}}\right)^3\\
j_{30B}(\tau) &= j_{30D}(\tau)+\frac{1}{ j_{30D}(\tau)}+2\\
j_{30C}(\tau) &= \left(\frac{d_1\, d_3\, d_{5}\, d_{15}}{d_2\, d_{6}\, d_{10}\, d_{30}}\right)\\
j_{30D}(\tau) &= \left(\frac{d_2\, d_3\, d_{10}\, d_{15}}{d_1\, d_{5}\, d_{6}\, d_{30}}\right)^2\\
j_{30E}(\tau) &= \left(\frac{d_{6}\, d_{15}}{d_{3}\, d_{30}}\right)^2\\
j_{30F}(\tau) &= \left(\frac{d_3\, d_5\, d_{6}\, d_{10}}{d_1\, d_{2}\, d_{15}\, d_{30}}\right)\\
j_{30G}(\tau) &= \left(\frac{d_1^2\, d_6\, d_{10}\, d_{15}^2}{d_2^2\, d_{3}\, d_{5}\, d_{30}^2}\right)\end{align}$$
These functions obey the triple relations. Let \(a=\color{red}{j_{30G}},\;m=1,\;n=2\), then,
$$\begin{align}
j_{30F} &= a+\frac{mn}{a}+(m+n)\\
j_{30C} &= a+m+\frac{m(m-n)}{a+m}-(2m-n)\\
j_{30A} &= a+n+\frac{n(m-n)}{a+n}+(m-2n)
\end{align}$$ as well as,
$$j_{30B} = j_{30F}+\frac{(m-n)^2}{j_{30F}}+3 = j_{30C}+\frac{n^2}{j_{30C}}+5 = j_{30A}+\frac{m^2}{j_{30A}}+7$$ which implies the linear relation between 5 monster functions,
$$j_{30B}+2\color{red}{j_{30G}} = j_{30A}+j_{30C}+j_{30F}+3$$ This is the highest level with linear relations between 5 monster functions.
j_{30A}(\tau) &= \left(\frac{d_1\, d_6\, d_{10}\, d_{15}}{d_2\, d_{3}\, d_{5}\, d_{30}}\right)^3\\
j_{30B}(\tau) &= j_{30D}(\tau)+\frac{1}{ j_{30D}(\tau)}+2\\
j_{30C}(\tau) &= \left(\frac{d_1\, d_3\, d_{5}\, d_{15}}{d_2\, d_{6}\, d_{10}\, d_{30}}\right)\\
j_{30D}(\tau) &= \left(\frac{d_2\, d_3\, d_{10}\, d_{15}}{d_1\, d_{5}\, d_{6}\, d_{30}}\right)^2\\
j_{30E}(\tau) &= \left(\frac{d_{6}\, d_{15}}{d_{3}\, d_{30}}\right)^2\\
j_{30F}(\tau) &= \left(\frac{d_3\, d_5\, d_{6}\, d_{10}}{d_1\, d_{2}\, d_{15}\, d_{30}}\right)\\
j_{30G}(\tau) &= \left(\frac{d_1^2\, d_6\, d_{10}\, d_{15}^2}{d_2^2\, d_{3}\, d_{5}\, d_{30}^2}\right)\end{align}$$
These functions obey the triple relations. Let \(a=\color{red}{j_{30G}},\;m=1,\;n=2\), then,
$$\begin{align}
j_{30F} &= a+\frac{mn}{a}+(m+n)\\
j_{30C} &= a+m+\frac{m(m-n)}{a+m}-(2m-n)\\
j_{30A} &= a+n+\frac{n(m-n)}{a+n}+(m-2n)
\end{align}$$ as well as,
$$j_{30B} = j_{30F}+\frac{(m-n)^2}{j_{30F}}+3 = j_{30C}+\frac{n^2}{j_{30C}}+5 = j_{30A}+\frac{m^2}{j_{30A}}+7$$ which implies the linear relation between 5 monster functions,
$$j_{30B}+2\color{red}{j_{30G}} = j_{30A}+j_{30C}+j_{30F}+3$$ This is the highest level with linear relations between 5 monster functions.
Level 24, part 1
Level 24
$$\begin{align}
j_{24A}(\tau) &= \left(\frac{d_4^2\, d_{12}^2}{d_2\, d_6\, d_8\, d_{24}}\right)^3=\sqrt{j_{12A}(2\tau)}\\
&=j_{24D}(\tau)+\frac{4}{j_{24D}(\tau)}\\
&=j_{24H}(\tau)+\frac{1}{j_{24H}(\tau)}\\
\color{blue}{j_{24B}(\tau)} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2\\
\color{red}{j_{24C}(\tau)} &= \left(\frac{d_1^2\, d_6\, d_8^2\, d_{12}}{d_2\, d_3^2\, d_4\, d_{24}^2}\right)\\
j_{24D}(\tau) &= \left(\frac{d_2\, d_6}{d_8\, d_{24}}\right)=\sqrt{j_{12E}(2\tau)}\\
j_{24E}(\tau) &= \left(\frac{d_{12}^2}{d_6\, d_{24}}\right)^4=\sqrt{j_{12D}(2\tau)}\\
j_{24F}(\tau) &= \left(\frac{d_8\, d_{12}}{d_4\, d_{24}}\right)^3 = \sqrt{j_{12F}(2\tau)}\\
j_{24G}(\tau) &= \left(\frac{d_4\, d_8}{d_{12}\, d_{24}}\right) = \sqrt{j_{12G}(2\tau)}\\
j_{24H}(\tau) &= \left(\frac{d_6\, d_8}{d_2\, d_{24}}\right)^2=\sqrt{j_{12H}(2\tau)}\\
\color{red}{j_{24I}(\tau)} &= \left(\frac{d_2\, d_3^2\, d_8^2\, d_{12}}{d_1^2\, d_4\, d_6\, d_{24}^2}\right)\\
j_{24J}(\tau) &= \;\left(\frac{d_{12}}{d_{24}}\right)^2\;=\;\sqrt{j_{12J}(2\tau)}\\
\end{align}$$ The "important" functions (that aren't just square roots) are \(j_{24C}\) and \(j_{24I}\). First define the auxiliary non-monster functions, $$\begin{align}
U &=\left(\frac{d_1\,d_6\,d_8\,d_{12}}{d_2\,d_3\,d_4\,d_{24}}\right)^4\\
V &=\left(\frac{d_2^2\,d_3\,d_8\,d_{12}^2}{d_1\,d_4^2\,d_6^2\,d_{24}}\right)^4\quad\quad
\end{align}$$ I. Let \(a_1=\color{red}{j_{24C}},\;m = 1,\;n=3,\) then,
$$\begin{align}
\color{blue}{j_{24B}} &= a_1+\frac{mn}{a_1}+(m+n)\\
j_{12G} &= a_1+m+\frac{m(m-n)}{a_1+m}-(2m-n)\\
U &= a_1+n+\frac{n(m-n)}{a_1+n}+(m-2n)
\end{align}$$ as well as, $$j_{12C} = j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 = j_{12G}+\frac{n^2}{j_{12G}} = U+\frac{m^2}{U}+4$$
II. Let \(a_2=\color{red}{j_{24I}},\;m = 1,\;n=-1,\) then,
$$\begin{align}
\color{blue}{j_{24B}} &= a_2+\frac{mn}{a_2}+(m+n)\\
j_{12F} &= a_2+m+\frac{m(m-n)}{a_2+m}-(2m-n)\\
V &= a_2+n+\frac{n(m-n)}{a_2+n}+(m-2n)
\end{align}$$ as well as, $$j_{12C} = j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 = j_{12F}+\frac{n^2}{j_{12F}} = V+\frac{m^2}{V}-4$$ These two families imply the two linear relations, $$j_{12C}+2\color{red}{j_{24C}} = j_{24B}+j_{12G}+U-2\\j_{12C}+2\color{red}{j_{24I}} = j_{24B}+j_{12F}+V-2$$
However, there is one linear relation using only monster functions,
$$j_{12C}-j_{12F}-j_{12G}+j_{12E}-2j_{12I} = 2\big(\color{blue}{j_{24B}}-\color{red}{j_{24C}}-\color{red}{j_{24I}}-3\big)$$ Note that,
$$j_{12C}=j_{12F}+\frac{1}{j_{12F}} = j_{12G}+\frac{9}{j_{12G}}\\ j_{12E} = j_{12I}-\frac{3}{j_{12I}}-2\quad$$
$$\begin{align}
j_{24A}(\tau) &= \left(\frac{d_4^2\, d_{12}^2}{d_2\, d_6\, d_8\, d_{24}}\right)^3=\sqrt{j_{12A}(2\tau)}\\
&=j_{24D}(\tau)+\frac{4}{j_{24D}(\tau)}\\
&=j_{24H}(\tau)+\frac{1}{j_{24H}(\tau)}\\
\color{blue}{j_{24B}(\tau)} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2\\
\color{red}{j_{24C}(\tau)} &= \left(\frac{d_1^2\, d_6\, d_8^2\, d_{12}}{d_2\, d_3^2\, d_4\, d_{24}^2}\right)\\
j_{24D}(\tau) &= \left(\frac{d_2\, d_6}{d_8\, d_{24}}\right)=\sqrt{j_{12E}(2\tau)}\\
j_{24E}(\tau) &= \left(\frac{d_{12}^2}{d_6\, d_{24}}\right)^4=\sqrt{j_{12D}(2\tau)}\\
j_{24F}(\tau) &= \left(\frac{d_8\, d_{12}}{d_4\, d_{24}}\right)^3 = \sqrt{j_{12F}(2\tau)}\\
j_{24G}(\tau) &= \left(\frac{d_4\, d_8}{d_{12}\, d_{24}}\right) = \sqrt{j_{12G}(2\tau)}\\
j_{24H}(\tau) &= \left(\frac{d_6\, d_8}{d_2\, d_{24}}\right)^2=\sqrt{j_{12H}(2\tau)}\\
\color{red}{j_{24I}(\tau)} &= \left(\frac{d_2\, d_3^2\, d_8^2\, d_{12}}{d_1^2\, d_4\, d_6\, d_{24}^2}\right)\\
j_{24J}(\tau) &= \;\left(\frac{d_{12}}{d_{24}}\right)^2\;=\;\sqrt{j_{12J}(2\tau)}\\
\end{align}$$ The "important" functions (that aren't just square roots) are \(j_{24C}\) and \(j_{24I}\). First define the auxiliary non-monster functions, $$\begin{align}
U &=\left(\frac{d_1\,d_6\,d_8\,d_{12}}{d_2\,d_3\,d_4\,d_{24}}\right)^4\\
V &=\left(\frac{d_2^2\,d_3\,d_8\,d_{12}^2}{d_1\,d_4^2\,d_6^2\,d_{24}}\right)^4\quad\quad
\end{align}$$ I. Let \(a_1=\color{red}{j_{24C}},\;m = 1,\;n=3,\) then,
$$\begin{align}
\color{blue}{j_{24B}} &= a_1+\frac{mn}{a_1}+(m+n)\\
j_{12G} &= a_1+m+\frac{m(m-n)}{a_1+m}-(2m-n)\\
U &= a_1+n+\frac{n(m-n)}{a_1+n}+(m-2n)
\end{align}$$ as well as, $$j_{12C} = j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 = j_{12G}+\frac{n^2}{j_{12G}} = U+\frac{m^2}{U}+4$$
II. Let \(a_2=\color{red}{j_{24I}},\;m = 1,\;n=-1,\) then,
$$\begin{align}
\color{blue}{j_{24B}} &= a_2+\frac{mn}{a_2}+(m+n)\\
j_{12F} &= a_2+m+\frac{m(m-n)}{a_2+m}-(2m-n)\\
V &= a_2+n+\frac{n(m-n)}{a_2+n}+(m-2n)
\end{align}$$ as well as, $$j_{12C} = j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 = j_{12F}+\frac{n^2}{j_{12F}} = V+\frac{m^2}{V}-4$$ These two families imply the two linear relations, $$j_{12C}+2\color{red}{j_{24C}} = j_{24B}+j_{12G}+U-2\\j_{12C}+2\color{red}{j_{24I}} = j_{24B}+j_{12F}+V-2$$
However, there is one linear relation using only monster functions,
$$j_{12C}-j_{12F}-j_{12G}+j_{12E}-2j_{12I} = 2\big(\color{blue}{j_{24B}}-\color{red}{j_{24C}}-\color{red}{j_{24I}}-3\big)$$ Note that,
$$j_{12C}=j_{12F}+\frac{1}{j_{12F}} = j_{12G}+\frac{9}{j_{12G}}\\ j_{12E} = j_{12I}-\frac{3}{j_{12I}}-2\quad$$
Level 20
I. Monster functions: Let,
$$a(\tau) = \left(\frac{d_1\, d_4\, d_{10}}{d_2\, d_5\, d_{20}}\right)^{2},\quad b(\tau) = \frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}, \quad c(\tau) = \frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}$$
such that \(a+1=b,\; a+5=c\). This level 20 triple can be derived from a level 10 triple. Then define,
$$\begin{align}
j_{20A}(\tau) &= \left(\frac{d_2^2\, d_{10}^2}{d_1\, d_4\, d_5\, d_{20}}\right)^4=\frac{bc}{a} \\
&= \left( \sqrt{j_{20F}}+\frac1{\sqrt{ j_{20F}}}\right)^2\\
j_{20B}(\tau) &=\sqrt{j_{10A}(2\tau)}\\
&= j_{20D}(\tau)+\frac4{j_{20D}(\tau)}\\
&= j_{20E}(\tau)-\frac1{j_{20E}(\tau)} \\
\color{red}{j_{20C}}(\tau) &= \left(\frac{d_1\, d_4\, d_{10}}{d_2\, d_5\, d_{20}}\right)^{2}\;\;=\;\;a\\
j_{20D}(\tau) &= \left(\frac{d_2\, d_{10}}{d_4\, d_{20}}\right)^{2}=\sqrt{j_{10B}(2\tau)}\\
j_{20E}(\tau) &= \left(\frac{d_4\, d_{10}}{d_2\, d_{20}}\right)^{3}=\sqrt{j_{10D}(2\tau)}\\
j_{20F}(\tau) &= \left(\frac{d_4\, d_5}{d_1\, d_{20}}\right)^{2}
\end{align}$$
Examples: \(\; j_{10A}\left(\tfrac12\sqrt{-19/10}\right) = 76^2,\;\; j_{20B}\left(\tfrac14\sqrt{-19/10}\right) = 76\)
II. Non-monster functions. Also define,
$$\begin{align}
u(\tau) &= \left(\frac{d_2^4\,d_5\,d_{20}}{d_1\,d_4\,d_{10}^4}\right)^2 = \frac{ac}b\\
v(\tau) &= \left(\frac{d_1\,d_4\,d_{10}^2}{d_2^2\,d_5\,d_{20}}\right)^6 =\frac{ab}c
\end{align}$$
III. Relations. Like \(j_{30B}\), the function \(j_{20A}\) is expressible by others in 4 ways. There is no linear relation between purely monster functions of level 20.
$$a(\tau) = \left(\frac{d_1\, d_4\, d_{10}}{d_2\, d_5\, d_{20}}\right)^{2},\quad b(\tau) = \frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}, \quad c(\tau) = \frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}$$
such that \(a+1=b,\; a+5=c\). This level 20 triple can be derived from a level 10 triple. Then define,
$$\begin{align}
j_{20A}(\tau) &= \left(\frac{d_2^2\, d_{10}^2}{d_1\, d_4\, d_5\, d_{20}}\right)^4=\frac{bc}{a} \\
&= \left( \sqrt{j_{20F}}+\frac1{\sqrt{ j_{20F}}}\right)^2\\
j_{20B}(\tau) &=\sqrt{j_{10A}(2\tau)}\\
&= j_{20D}(\tau)+\frac4{j_{20D}(\tau)}\\
&= j_{20E}(\tau)-\frac1{j_{20E}(\tau)} \\
\color{red}{j_{20C}}(\tau) &= \left(\frac{d_1\, d_4\, d_{10}}{d_2\, d_5\, d_{20}}\right)^{2}\;\;=\;\;a\\
j_{20D}(\tau) &= \left(\frac{d_2\, d_{10}}{d_4\, d_{20}}\right)^{2}=\sqrt{j_{10B}(2\tau)}\\
j_{20E}(\tau) &= \left(\frac{d_4\, d_{10}}{d_2\, d_{20}}\right)^{3}=\sqrt{j_{10D}(2\tau)}\\
j_{20F}(\tau) &= \left(\frac{d_4\, d_5}{d_1\, d_{20}}\right)^{2}
\end{align}$$
Examples: \(\; j_{10A}\left(\tfrac12\sqrt{-19/10}\right) = 76^2,\;\; j_{20B}\left(\tfrac14\sqrt{-19/10}\right) = 76\)
II. Non-monster functions. Also define,
$$\begin{align}
u(\tau) &= \left(\frac{d_2^4\,d_5\,d_{20}}{d_1\,d_4\,d_{10}^4}\right)^2 = \frac{ac}b\\
v(\tau) &= \left(\frac{d_1\,d_4\,d_{10}^2}{d_2^2\,d_5\,d_{20}}\right)^6 =\frac{ab}c
\end{align}$$
III. Relations. Like \(j_{30B}\), the function \(j_{20A}\) is expressible by others in 4 ways. There is no linear relation between purely monster functions of level 20.
Level 18
This level is also special since we can find two \(a(\tau)\) namely, \(\color{red}{j_{18D}}\) and \(\color{blue}{j_{18C}}\). Using S1 of the Intro, these two explain \(2\times4 = 8\) of the 14 trinomial identities of level 18 found by Somos.
I. First triple. Define,
$$a(\tau) =\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right),\quad b(\tau) = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right),\quad c(\tau) =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ then,
$$\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-1 = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right)\quad \\ \left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-3 =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ so \(a-1 = b,\,a-3 = c,\) and we get the moonshine functions,
$$\begin{align}
j_{18A}(\tau) &= \left(\frac{d_1\, d_2}{d_9\, d_{18}}\right) = \frac{ac}{b}\\
j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\\
\color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\, d_9}{d_2^2\,d_3^2 d_{18}^2}\right)^2 =\frac{bc}{a}+1\\
\color{red}{j_{18D}}(\tau) &=\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)\;=\; a \\
j_{18E}(\tau) &= \left(\frac{d_2\, d_9}{d_1\, d_{18}}\right)^{3}= \frac{ab}{c}\\
\end{align}$$ Naturally, these also satisfy S2 and S3. We also have
$$j_{18B} = j_{18A}+\frac{3^2}{j_{18A}}+3 = j_{18E}+\frac{1}{j_{18E}}-1$$ and the linear relation between these 5 moonshine functions,
$$j_{18B}+2\color{red}{j_{18D}} = j_{18A}+j_{18C}+j_{18E}+4$$
II. Second triple. Define,
$$a(\tau) =\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2,\quad b(\tau) = \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right),\quad c(\tau) =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$ then, $$\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2-1 = \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right)\quad\\ \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2+3 =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$so \(a-1 = b,\,a+3 = c,\) and we get,
$$\begin{align}
j_{6d} &= j_{6F}+\frac{16}{j_{6F}}\\
j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2} =\frac{ac}{b}\\
\color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2 \;\;=\;\;a\\
\beta(\tau) &=\left(\frac{d_1\, d_6^2\, d_9}{d_2\, d_3^2\, d_{18}}\right)^6 =\frac{ab}{c} \\
j_{6F}(\tau) &=\; \left(\frac{d_3}{d_6}\right)^8 \;\;= \;\;\frac{bc}{a}\\
\end{align}$$ where \(j_{6d}\) and \(\beta\) are non-monster functions with expansions (A007263) and (A128512), respectively. Naturally, these also satisfy S1,2,3. And by adding these additional relations, we get a second linear equation for level 18 of similar form to the first,
$$j_{6d}+2\color{blue}{j_{18C}}=j_{18B}+j_{6F}+\beta-2$$
I. First triple. Define,
$$a(\tau) =\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right),\quad b(\tau) = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right),\quad c(\tau) =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ then,
$$\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-1 = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right)\quad \\ \left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-3 =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ so \(a-1 = b,\,a-3 = c,\) and we get the moonshine functions,
$$\begin{align}
j_{18A}(\tau) &= \left(\frac{d_1\, d_2}{d_9\, d_{18}}\right) = \frac{ac}{b}\\
j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\\
\color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\, d_9}{d_2^2\,d_3^2 d_{18}^2}\right)^2 =\frac{bc}{a}+1\\
\color{red}{j_{18D}}(\tau) &=\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)\;=\; a \\
j_{18E}(\tau) &= \left(\frac{d_2\, d_9}{d_1\, d_{18}}\right)^{3}= \frac{ab}{c}\\
\end{align}$$ Naturally, these also satisfy S2 and S3. We also have
$$j_{18B} = j_{18A}+\frac{3^2}{j_{18A}}+3 = j_{18E}+\frac{1}{j_{18E}}-1$$ and the linear relation between these 5 moonshine functions,
$$j_{18B}+2\color{red}{j_{18D}} = j_{18A}+j_{18C}+j_{18E}+4$$
II. Second triple. Define,
$$a(\tau) =\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2,\quad b(\tau) = \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right),\quad c(\tau) =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$ then, $$\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2-1 = \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right)\quad\\ \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2+3 =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$so \(a-1 = b,\,a+3 = c,\) and we get,
$$\begin{align}
j_{6d} &= j_{6F}+\frac{16}{j_{6F}}\\
j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2} =\frac{ac}{b}\\
\color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2 \;\;=\;\;a\\
\beta(\tau) &=\left(\frac{d_1\, d_6^2\, d_9}{d_2\, d_3^2\, d_{18}}\right)^6 =\frac{ab}{c} \\
j_{6F}(\tau) &=\; \left(\frac{d_3}{d_6}\right)^8 \;\;= \;\;\frac{bc}{a}\\
\end{align}$$ where \(j_{6d}\) and \(\beta\) are non-monster functions with expansions (A007263) and (A128512), respectively. Naturally, these also satisfy S1,2,3. And by adding these additional relations, we get a second linear equation for level 18 of similar form to the first,
$$j_{6d}+2\color{blue}{j_{18C}}=j_{18B}+j_{6F}+\beta-2$$
Level 16
I. Moonshine functions: Define,$$a(\tau)= \left(\frac{d_8^3}{d_4\, d_{16}^2}\right)^{2}, \quad b(\tau) = \left(\frac{d_1^2\,d_8}{d_2\,d_{16}^2}\right), \quad c(\tau) = \left(\frac{d_2^5\,d_8}{d_1^2\,d_4^2\,d_{16}^2}\right)$$ then, $$\left(\frac{d_8^3}{d_4\, d_{16}^2}\right)^{2}-2= \left(\frac{d_1^2\,d_8}{d_2\,d_{16}^2}\right)\quad\\ \left(\frac{d_8^3}{d_4\, d_{16}^2}\right)^{2}+2 = \left(\frac{d_2^5\,d_8}{d_1^2\,d_4^2\,d_{16}^2}\right)$$so \(a-2 = b,\, a+2 = c\). This \(a,b,c\) also belong to the class where \(a(\tau) = -a\big(\tfrac12+\tau\big)\) and \(b(\tau) = -c\big(\tfrac12+\tau\big)\). And the three level 16 moonshine functions,$$\begin{align}
j_{16A}(\tau) &= \left(\frac{d_4\, d_8}{d_2\, d_{16}}\right)^{4}=\sqrt{j_{8A}(2\tau)}\\
\color{red}{j_{16B}}(\tau) &= \left(\frac{d_8^3}{d_4\, d_{16}^2}\right)^{2}\; = \; a\\
j_{16C}(\tau) &= \left(\frac{d_2^3\,d_8^3}{d_1^2\, d_4^2\, d_{16}^2}\right)^{2} = \frac{ac}{b}
\end{align}$$
II. Relations: Using S1, this \(a,b,c\) explains 4 of the 6 trinomial identities of level 16 (prefixed with t16) by Somos. They also obey,
$$\left(\frac{bc}{a}+16\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ where \(b-c=4\). Or in terms of named functions of level 8 and 16, $$\quad\quad j_{8B}\;+\;2\color{red}{j_{16B}} = j_{8D}+j_{16C}+j_{16C'}$$where, similar to level 8, $$j_{16C'}(\tau) =-j_{16C}\big(\tfrac12+\tau\big)= \;\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}$$ The last 2 of the 6 trinomial identities involves the pair \(j_{16C}\) and \(j_{16C'}\), hence is just the same function using different arguments, $$\begin{align}j_{16C}(\tau)-2 &=\left(\frac{d_2^3\,d_8^3}{d_1^2\, d_4^2\, d_{16}^2}\right)^{2}-2 =\frac{d_4^{10}}{d_1^2\,d_2^3\,d_8^3\,d_{16}^2}\\ j_{16C}\big(\tfrac12+\tau\big)-2 &= \; -\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}-2 \;= \; -\frac{d_1^2\,d_4^{12}}{d_2^9\,d_8^3\,d_{16}^2}\end{align}$$
III. Inter-level quadratic relations: We also have,
$$\begin{align}
j_{4C} &= j_{8E}+\frac{16}{j_{8E}}\\
j_{4D} &= j_{8E}-\frac{16}{j_{8E}}
\end{align}$$ $$\begin{align}j_{8D} &= j_{16B}-\frac{4}{j_{16B}}\\
j_{8E} &= j_{16B}+\frac{4}{j_{16B}}\quad\end{align}$$ Adding each pair leaves functions with the same argument,
$$\begin{align}
j_{4C}+j_{4D} &= 2\color{red}{j_{8E}}\\
j_{8D}+j_{8E} &= 2\color{red}{j_{16B}}
\end{align}$$ which are 2 of the 9 linear relations found by Conway, Norton, and Atkins.
j_{16A}(\tau) &= \left(\frac{d_4\, d_8}{d_2\, d_{16}}\right)^{4}=\sqrt{j_{8A}(2\tau)}\\
\color{red}{j_{16B}}(\tau) &= \left(\frac{d_8^3}{d_4\, d_{16}^2}\right)^{2}\; = \; a\\
j_{16C}(\tau) &= \left(\frac{d_2^3\,d_8^3}{d_1^2\, d_4^2\, d_{16}^2}\right)^{2} = \frac{ac}{b}
\end{align}$$
II. Relations: Using S1, this \(a,b,c\) explains 4 of the 6 trinomial identities of level 16 (prefixed with t16) by Somos. They also obey,
$$\left(\frac{bc}{a}+16\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ where \(b-c=4\). Or in terms of named functions of level 8 and 16, $$\quad\quad j_{8B}\;+\;2\color{red}{j_{16B}} = j_{8D}+j_{16C}+j_{16C'}$$where, similar to level 8, $$j_{16C'}(\tau) =-j_{16C}\big(\tfrac12+\tau\big)= \;\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}$$ The last 2 of the 6 trinomial identities involves the pair \(j_{16C}\) and \(j_{16C'}\), hence is just the same function using different arguments, $$\begin{align}j_{16C}(\tau)-2 &=\left(\frac{d_2^3\,d_8^3}{d_1^2\, d_4^2\, d_{16}^2}\right)^{2}-2 =\frac{d_4^{10}}{d_1^2\,d_2^3\,d_8^3\,d_{16}^2}\\ j_{16C}\big(\tfrac12+\tau\big)-2 &= \; -\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}-2 \;= \; -\frac{d_1^2\,d_4^{12}}{d_2^9\,d_8^3\,d_{16}^2}\end{align}$$
III. Inter-level quadratic relations: We also have,
$$\begin{align}
j_{4C} &= j_{8E}+\frac{16}{j_{8E}}\\
j_{4D} &= j_{8E}-\frac{16}{j_{8E}}
\end{align}$$ $$\begin{align}j_{8D} &= j_{16B}-\frac{4}{j_{16B}}\\
j_{8E} &= j_{16B}+\frac{4}{j_{16B}}\quad\end{align}$$ Adding each pair leaves functions with the same argument,
$$\begin{align}
j_{4C}+j_{4D} &= 2\color{red}{j_{8E}}\\
j_{8D}+j_{8E} &= 2\color{red}{j_{16B}}
\end{align}$$ which are 2 of the 9 linear relations found by Conway, Norton, and Atkins.
Level 12, part 1
Level 12 is very special because, given an eta quotient \(a\), we can find \(a+m_i\) as an eta quotient for four different integer values \(m_i\). However, that also makes the situation more complicated than in other levels. Let,
$$a(\tau) = \left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2,\quad b(\tau) = \frac{d_3^3\,d_4}{d_1\,d_{12}^3}, \quad c(\tau) =\frac{d_1^3\,d_4\, d_6^2}{d_2^2\,d_3\,d_{12}^3}$$ such that \(a+1 = b,\,a-3 = c.\) And the monster functions, $$\begin{align}
j_{12A}(\tau) &= \left(\frac{d_2^2\, d_6^2}{d_1\,d_3\,d_4\, d_{12}}\right)^{6}\\
\color{red}{j_{12B}}(\tau) &= \left(\frac{d_1\, d_4\,d_6}{d_2\,d_3\, d_{12}}\right)^4\,=\,\frac{ac}b\\
j_{12C}(\tau) &= \sqrt{j_{6A}(2\tau)}\\
&=j_{12F}(\tau)+\frac{1}{j_{12F}(\tau)}\\
&= j_{12G}(\tau)+\frac{9}{j_{12G}(\tau)}\\
j_{12D}(\tau) &= \left(\frac{d_6^2}{d_3\, d_{12}}\right)^{8}=\sqrt[3]{j_{4A}(3\tau)}\\
\color{blue}{j_{12E}}(\tau) &= \left(\frac{d_1\, d_3}{d_4\, d_{12}}\right)^{2}\;=\;\;\frac{bc}a\\
j_{12F}(\tau) &= \left(\frac{d_4\, d_6}{d_2\, d_{12}}\right)^{6}=\sqrt{j_{6B}(2\tau)}\\
j_{12G}(\tau) &= \left(\frac{d_2\, d_4}{d_6\, d_{12}}\right)^{2}=\sqrt{j_{6D}(2\tau)}\\
\color{blue}{j_{12H}}(\tau) &= \left(\frac{d_3\, d_4}{d_1\, d_{12}}\right)^{4}\;=\;\frac{ab}c\\
\color{red}{j_{12I}}(\tau) &= \left(\frac{d_4^2\, d_6}{d_2\, d_{12}^2}\right)^{2}\;=\; a\\
j_{12J}(\tau) &= \;\left(\frac{d_6}{d_{12}}\right)^{4}\; = \;\sqrt{j_{6F}(2\tau)}
\end{align}$$
Example: \(\quad j_{12C}\left(\tfrac12\sqrt{-17/6}\right) = 198\)
Let \(a=\color{red}{j_{12I}},\;m = 1,\;n=-3,\) then,
$$\begin{align}
j_{12E} &= a-\frac{3}{a}-2\\
j_{12B} &= a+1+\frac{4}{a+1}-5\\
j_{12H} &= a-3+\frac{12}{a-3}+7
\end{align}$$ as well as,
$$j_{12A} = j_{12E}+\frac{4^2}{j_{12E}}+8 = j_{12B}+\frac{3^2}{j_{12B}}+10 = j_{12H}+\frac{1}{j_{12H}}+2$$ which implies the linear relation between 5 monster functions,
$$j_{12A}+2\color{red}{j_{12I}} =\color{red}{ j_{12B}}+\color{blue}{j_{12E}}+\color{blue}{j_{12H}}+8$$ The functions \(j_{12I}\) and \(j_{12B}\) can be used for level 12, while versions of \(j_{12E}\) and \(j_{12H}\) will be useful in level 24.
$$a(\tau) = \left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2,\quad b(\tau) = \frac{d_3^3\,d_4}{d_1\,d_{12}^3}, \quad c(\tau) =\frac{d_1^3\,d_4\, d_6^2}{d_2^2\,d_3\,d_{12}^3}$$ such that \(a+1 = b,\,a-3 = c.\) And the monster functions, $$\begin{align}
j_{12A}(\tau) &= \left(\frac{d_2^2\, d_6^2}{d_1\,d_3\,d_4\, d_{12}}\right)^{6}\\
\color{red}{j_{12B}}(\tau) &= \left(\frac{d_1\, d_4\,d_6}{d_2\,d_3\, d_{12}}\right)^4\,=\,\frac{ac}b\\
j_{12C}(\tau) &= \sqrt{j_{6A}(2\tau)}\\
&=j_{12F}(\tau)+\frac{1}{j_{12F}(\tau)}\\
&= j_{12G}(\tau)+\frac{9}{j_{12G}(\tau)}\\
j_{12D}(\tau) &= \left(\frac{d_6^2}{d_3\, d_{12}}\right)^{8}=\sqrt[3]{j_{4A}(3\tau)}\\
\color{blue}{j_{12E}}(\tau) &= \left(\frac{d_1\, d_3}{d_4\, d_{12}}\right)^{2}\;=\;\;\frac{bc}a\\
j_{12F}(\tau) &= \left(\frac{d_4\, d_6}{d_2\, d_{12}}\right)^{6}=\sqrt{j_{6B}(2\tau)}\\
j_{12G}(\tau) &= \left(\frac{d_2\, d_4}{d_6\, d_{12}}\right)^{2}=\sqrt{j_{6D}(2\tau)}\\
\color{blue}{j_{12H}}(\tau) &= \left(\frac{d_3\, d_4}{d_1\, d_{12}}\right)^{4}\;=\;\frac{ab}c\\
\color{red}{j_{12I}}(\tau) &= \left(\frac{d_4^2\, d_6}{d_2\, d_{12}^2}\right)^{2}\;=\; a\\
j_{12J}(\tau) &= \;\left(\frac{d_6}{d_{12}}\right)^{4}\; = \;\sqrt{j_{6F}(2\tau)}
\end{align}$$
Example: \(\quad j_{12C}\left(\tfrac12\sqrt{-17/6}\right) = 198\)
Let \(a=\color{red}{j_{12I}},\;m = 1,\;n=-3,\) then,
$$\begin{align}
j_{12E} &= a-\frac{3}{a}-2\\
j_{12B} &= a+1+\frac{4}{a+1}-5\\
j_{12H} &= a-3+\frac{12}{a-3}+7
\end{align}$$ as well as,
$$j_{12A} = j_{12E}+\frac{4^2}{j_{12E}}+8 = j_{12B}+\frac{3^2}{j_{12B}}+10 = j_{12H}+\frac{1}{j_{12H}}+2$$ which implies the linear relation between 5 monster functions,
$$j_{12A}+2\color{red}{j_{12I}} =\color{red}{ j_{12B}}+\color{blue}{j_{12E}}+\color{blue}{j_{12H}}+8$$ The functions \(j_{12I}\) and \(j_{12B}\) can be used for level 12, while versions of \(j_{12E}\) and \(j_{12H}\) will be useful in level 24.
Level 10
I. Moonshine functions: Define, $$a(\tau) =\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2,\quad b(\tau) = \left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right),\quad c(\tau) =\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ then $$\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-1=\left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right)\\ \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-5=\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ so \(a-1 = b,\,a-5 = c\) and we get the moonshine functions $$\begin{align}
j_{10A}(\tau) &= j_{10B}(\tau)+\frac{4^2}{j_{10B}(\tau)}+8\\
j_{10B}(\tau) &= \left(\frac{d_1\, d_5}{d_2\, d_{10}}\right)^{4}=\frac{bc}{a}\\
j_{10C}(\tau) &= \left(\frac{d_1\, d_2}{d_5\, d_{10}}\right)^{2} =\frac{ac}{b}\\
j_{10D}(\tau) &= \left(\frac{d_2\, d_5}{d_1\, d_{10}}\right)^{6} = \frac{ab}{c}\\
j_{10E}(\tau) &= \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2 \;=\; a \\
\end{align}$$ Example: \(\quad j_{10A}\left(\sqrt{-19/10}\right) = 76^2\)
II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-5,\) then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let \(a=\color{red}{j_{10E}}\) then, $$\begin{align}
j_{10B} &= a+\frac{5}{a}-6\\
j_{10C} &= a-1-\frac{4}{a-1}-3\\
j_{10D} &= a-5+\frac{20}{a-5}+9
\end{align}$$ as well as S3, $$j_{10A} = j_{10B}+\frac{4^2}{j_{10B}}+8 = j_{10C}+\frac{5^2}{j_{10C}}+6 = j_{10D}+\frac{1}{j_{10D}}-2$$ which implies the linear relation between 5 moonshine functions,
$$j_{10A}+2\color{red}{j_{10E}} = j_{10B}+j_{10C}+j_{10D}+8$$
III. Special property. As mentioned in the Intro, there are special triples {\(a',b',c'\)} of level \(N = 6, 10, 12, 18, 30\) that, using a common formula, can generate another triple {\(a',b',c'\)} of level \(2N\). For \(N = 10\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}}{d_2\,d_5\,d_{20}}\right)^2 = j_{20C}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+5=c'\).
j_{10A}(\tau) &= j_{10B}(\tau)+\frac{4^2}{j_{10B}(\tau)}+8\\
j_{10B}(\tau) &= \left(\frac{d_1\, d_5}{d_2\, d_{10}}\right)^{4}=\frac{bc}{a}\\
j_{10C}(\tau) &= \left(\frac{d_1\, d_2}{d_5\, d_{10}}\right)^{2} =\frac{ac}{b}\\
j_{10D}(\tau) &= \left(\frac{d_2\, d_5}{d_1\, d_{10}}\right)^{6} = \frac{ab}{c}\\
j_{10E}(\tau) &= \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2 \;=\; a \\
\end{align}$$ Example: \(\quad j_{10A}\left(\sqrt{-19/10}\right) = 76^2\)
II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-5,\) then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let \(a=\color{red}{j_{10E}}\) then, $$\begin{align}
j_{10B} &= a+\frac{5}{a}-6\\
j_{10C} &= a-1-\frac{4}{a-1}-3\\
j_{10D} &= a-5+\frac{20}{a-5}+9
\end{align}$$ as well as S3, $$j_{10A} = j_{10B}+\frac{4^2}{j_{10B}}+8 = j_{10C}+\frac{5^2}{j_{10C}}+6 = j_{10D}+\frac{1}{j_{10D}}-2$$ which implies the linear relation between 5 moonshine functions,
$$j_{10A}+2\color{red}{j_{10E}} = j_{10B}+j_{10C}+j_{10D}+8$$
III. Special property. As mentioned in the Intro, there are special triples {\(a',b',c'\)} of level \(N = 6, 10, 12, 18, 30\) that, using a common formula, can generate another triple {\(a',b',c'\)} of level \(2N\). For \(N = 10\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}}{d_2\,d_5\,d_{20}}\right)^2 = j_{20C}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+5=c'\).
Level 8
I. Moonshine functions: Define, $$a(\tau)=\left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4},\quad b(\tau) = \left(\frac{d_1^2\,d_4}{d_2\,d_8^2}\right)^2, \quad c(\tau) = \left(\frac{d_2^5}{d_1^2\,d_4\,d_8^2}\right)^2$$ then, $$ \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}-4 = \left(\frac{d_1^2\,d_4}{d_2\,d_8^2}\right)^2\quad\\ \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}+4=\left(\frac{d_2^5}{d_1^2\,d_4\,d_8^2}\right)^2$$so \(a-4 = b,\, a+4 = c\). This \(a,b,c\) also belong to the class where \(a(\tau) = -a\big(\tfrac12+\tau\big)\) and \(b(\tau) = -c\big(\tfrac12+\tau\big)\). We then have the six level 8 moonshine functions,
$$\qquad\begin{align}
j_{8A}(\tau) &= \left(\frac{d_2\, d_4}{d_1\, d_8}\right)^{8}\;=\;\frac{ac}b\\
j_{8B}(\tau) &= \left(\frac{d_4^2}{d_2\, d_8}\right)^{12}=\sqrt{j_{4A}(2\tau)}\\
j_{8C}(\tau) &= \sqrt{j_{4B}(\tau)}\,=\,\sqrt[4]{j_{2A}(2\tau)}\\
&= \sqrt{j_{8A}(\tau)}-\frac{4}{\sqrt{j_{8A}(\tau)}}\\
j_{8D}(\tau) &= \left(\frac{d_2}{d_8}\right)^{4}\\
\color{red}{j_{8E}}(\tau) &= \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}\;=\; a\\
j_{8F}(\tau) &= \left(\frac{d_4}{d_8}\right)^{6}
\end{align}$$ Example: \(\quad j_{8C}\big(\tfrac14\sqrt{-58}\big) = 396\)
II. Relations: Using S1, this \(a,b,c\) explains all of the four trinomial identities of level 8 (prefixed with t8) by Somos. They also obey the relation,
$$\left(\frac{bc}{a}+64\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ where \(b-c=8\). Or in terms of named functions of level 4 and 8,
$$\quad\quad j_{4B}\;+\;2\color{red}{j_{8E}} = j_{4D}+j_{8A}+j_{8A'}$$ where, $$j_{8A'}(\tau) = -j_{8A}\big(\tfrac12+\tau\big) = \left(\frac{d_1\, d_4^2}{d_2^2\, d_8}\right)^{8}$$
$$\qquad\begin{align}
j_{8A}(\tau) &= \left(\frac{d_2\, d_4}{d_1\, d_8}\right)^{8}\;=\;\frac{ac}b\\
j_{8B}(\tau) &= \left(\frac{d_4^2}{d_2\, d_8}\right)^{12}=\sqrt{j_{4A}(2\tau)}\\
j_{8C}(\tau) &= \sqrt{j_{4B}(\tau)}\,=\,\sqrt[4]{j_{2A}(2\tau)}\\
&= \sqrt{j_{8A}(\tau)}-\frac{4}{\sqrt{j_{8A}(\tau)}}\\
j_{8D}(\tau) &= \left(\frac{d_2}{d_8}\right)^{4}\\
\color{red}{j_{8E}}(\tau) &= \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}\;=\; a\\
j_{8F}(\tau) &= \left(\frac{d_4}{d_8}\right)^{6}
\end{align}$$ Example: \(\quad j_{8C}\big(\tfrac14\sqrt{-58}\big) = 396\)
II. Relations: Using S1, this \(a,b,c\) explains all of the four trinomial identities of level 8 (prefixed with t8) by Somos. They also obey the relation,
$$\left(\frac{bc}{a}+64\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ where \(b-c=8\). Or in terms of named functions of level 4 and 8,
$$\quad\quad j_{4B}\;+\;2\color{red}{j_{8E}} = j_{4D}+j_{8A}+j_{8A'}$$ where, $$j_{8A'}(\tau) = -j_{8A}\big(\tfrac12+\tau\big) = \left(\frac{d_1\, d_4^2}{d_2^2\, d_8}\right)^{8}$$
Level 6
I. Moonshine functions: Let, $$a(\tau)=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b(\tau)= \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c(\tau)=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ 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 we get the moonshine functions, $$\begin{align}
j_{6A}(\tau) &= j_{6D}(\tau)+\frac{9^2}{j_{6D}(\tau)}+18\\
j_{6B}(\tau) &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} \\
j_{6C}(\tau) &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} \\
j_{6D}(\tau) &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} \\
j_{6E}(\tau) &= \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4 = a\\
j_{6F}(\tau) &=\;\left(\frac{d_3}{d_6}\right)^{8}\\
\end{align}$$ Example: \(\; j_{6A}\left(\sqrt{-17/6}\right) = 198^2\)
II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-9,\) then the system S1 given in the Introduction,
$$\begin{align}
a+m &= b\\
a+n &= c\\
a(m-n)+bn &=cm\\
b-c &= (m-n)\end{align}$$ explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let \(a=\color{red}{j_{6E}}\) then,
$$\begin{align}
j_{6B}(\tau) &= a-9+\frac{72}{a-9}+17\\
j_{6C}(\tau) &= a+\frac{9}{a}-10\\
j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\
\end{align}$$ as well as S3, $$j_{6A} = j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18 $$ which implies the linear relation between 5 moonshine functions,
$$j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20$$These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.
III. Special property. As mentioned in the Intro, there are special triples {\(a',b',c'\)} of level \(N = 6, 10, 12, 18, 30\) that using a common formula can generate another triple {\(a',b',c'\)} of level \(2N\). For \(N = 6\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+9=c'\).
j_{6A}(\tau) &= j_{6D}(\tau)+\frac{9^2}{j_{6D}(\tau)}+18\\
j_{6B}(\tau) &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} \\
j_{6C}(\tau) &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} \\
j_{6D}(\tau) &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} \\
j_{6E}(\tau) &= \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4 = a\\
j_{6F}(\tau) &=\;\left(\frac{d_3}{d_6}\right)^{8}\\
\end{align}$$ Example: \(\; j_{6A}\left(\sqrt{-17/6}\right) = 198^2\)
II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-9,\) then the system S1 given in the Introduction,
$$\begin{align}
a+m &= b\\
a+n &= c\\
a(m-n)+bn &=cm\\
b-c &= (m-n)\end{align}$$ explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let \(a=\color{red}{j_{6E}}\) then,
$$\begin{align}
j_{6B}(\tau) &= a-9+\frac{72}{a-9}+17\\
j_{6C}(\tau) &= a+\frac{9}{a}-10\\
j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\
\end{align}$$ as well as S3, $$j_{6A} = j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18 $$ which implies the linear relation between 5 moonshine functions,
$$j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20$$These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.
III. Special property. As mentioned in the Intro, there are special triples {\(a',b',c'\)} of level \(N = 6, 10, 12, 18, 30\) that using a common formula can generate another triple {\(a',b',c'\)} of level \(2N\). For \(N = 6\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+9=c'\).
Level 4
I. Moonshine function Define, $$a(\tau) = \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8,\quad b(\tau) = \left(\frac{d_1}{d_4}\right)^8+8$$ then $$\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8=\left(\frac{d_1}{d_4}\right)^8+8$$ Equivalently, using the reciprocal of \(a(\tau)\), $$\left(\frac{\sqrt2\,d_1\,d_4^2}{d_2^3}\right)^8+\left(\frac{d_1^2\,d_4}{d_2^3}\right)^8 = 1$$ which are just versions of the single trinomial identity for level 4. We also have the four level 4 moonshine functions, $$\qquad\qquad\begin{align}
j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+16\left(\frac{d_4}{d_1}\right)^4\right)^2=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\\
j_{4B}(\tau) &=\left(\left(\frac{d_2\,d_4}{d_{1}\;d_8}\right)^4-4\left(\frac{d_{1}\;d_8}{d_2\,d_4}\right)^4\right)^2 = \sqrt{j_{2A}(2\tau)} \\
&=j_{4D}(\tau)+\frac{2^6}{j_{4D}(\tau)} \\
\color{red}{j_{4C}}(\tau) &= \left(\frac{d_1}{d_4}\right)^8+8\; = \;b(\tau)\\
j_{4D}(\tau) &= \left(\frac{d_2}{d_4}\right)^{12}
\end{align}$$ Note that the function \(j_{4C}(\tau) = \frac{16}{\lambda(2\tau)}-8=\left(\frac{d_1}{d_4}\right)^8+8\) with modular lambda function \(\lambda(\tau)\) is a normalized Hauptmodul.
II. Comparisons: Notice that,$$j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8,\quad j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2$$The last two will play analogous roles for levels 8 and 16. Also, \(j_{4A}\) is remarkably a 24th power. Versions appear in levels that are 4m divisors of 24, $$j_{4A}=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24},\quad j_{8B}=\left(\frac{d_4^2}{d_2\,d_8}\right)^{12},\quad j_{12D}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^{8},\quad j_{24E}=\left(\frac{d_{12}^2}{d_6\,d_{24}}\right)^{4}$$ Similarly, versions of \(j_{2B}\) appear in levels that are 2m divisors of 24, $$j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}$$
j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+16\left(\frac{d_4}{d_1}\right)^4\right)^2=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\\
j_{4B}(\tau) &=\left(\left(\frac{d_2\,d_4}{d_{1}\;d_8}\right)^4-4\left(\frac{d_{1}\;d_8}{d_2\,d_4}\right)^4\right)^2 = \sqrt{j_{2A}(2\tau)} \\
&=j_{4D}(\tau)+\frac{2^6}{j_{4D}(\tau)} \\
\color{red}{j_{4C}}(\tau) &= \left(\frac{d_1}{d_4}\right)^8+8\; = \;b(\tau)\\
j_{4D}(\tau) &= \left(\frac{d_2}{d_4}\right)^{12}
\end{align}$$ Note that the function \(j_{4C}(\tau) = \frac{16}{\lambda(2\tau)}-8=\left(\frac{d_1}{d_4}\right)^8+8\) with modular lambda function \(\lambda(\tau)\) is a normalized Hauptmodul.
II. Comparisons: Notice that,$$j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8,\quad j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2$$The last two will play analogous roles for levels 8 and 16. Also, \(j_{4A}\) is remarkably a 24th power. Versions appear in levels that are 4m divisors of 24, $$j_{4A}=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24},\quad j_{8B}=\left(\frac{d_4^2}{d_2\,d_8}\right)^{12},\quad j_{12D}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^{8},\quad j_{24E}=\left(\frac{d_{12}^2}{d_6\,d_{24}}\right)^{4}$$ Similarly, versions of \(j_{2B}\) appear in levels that are 2m divisors of 24, $$j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}$$
Levels 1, 2, 3
Recall our notation for the Dedekind eta function, $$\color{blue}{d_m = \eta(m\tau)}$$ where we suppress the argument \(\tau\) for ease of notation. Then we have the following moonshine functions:
Level 1
$$\begin{align}
j_{1A}(\tau) &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3
\end{align}$$ This is the well-known j-function \(j(\tau)\). In this form, it is easily seen to be a cube. Its most famous value is \(j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big)=-640320^3\) .
Level 2
$$\begin{align}
j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\
j_{2B}(\tau) &= \left(\frac{d_1}{d_2}\right)^{24}
\end{align}$$ Alternatively, $$j_{2A}(\tau) =\left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4$$ In this form, \(j_{2A}\) is seen to be a 4th power. The function doesn't have a name, but is notable for the value \(j_{2A}(\tfrac12\sqrt{-58})=396^4\) in Ramanujan's well-known pi formula. Note that,
$$j_{4B}(\tau) = \sqrt{j_{2A}(2\tau)}\\j_{8C}(\tau) = \sqrt[4]{j_{2A}(2\tau)}$$
Level 3
$$\begin{align}
j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\
j_{3B}(\tau) &= \left(\frac{d_1}{d_3}\right)^{12}\\
j_{3C}(\tau) &= \sqrt[3]{j_{1A}(3\tau)}
\end{align}$$ Alternatively, $$j_{3A}(\tau) = \left(\frac{d_1^2}{d_3^2}+\frac{9\,d_9^3}{d_1\,d_3^2}\right)^6$$ In this form, \(j_{3A}\) is seen to be also a cube. An example is \(j_{3A}\big(\tfrac{1+\sqrt{-89/3}}{2}\big)=-300^3\) which can also be used in a Ramanujan-type pi formula. Notice that \(j_{3C}\) is a cube root.
Level 1
$$\begin{align}
j_{1A}(\tau) &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3
\end{align}$$ This is the well-known j-function \(j(\tau)\). In this form, it is easily seen to be a cube. Its most famous value is \(j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big)=-640320^3\) .
Level 2
$$\begin{align}
j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\
j_{2B}(\tau) &= \left(\frac{d_1}{d_2}\right)^{24}
\end{align}$$ Alternatively, $$j_{2A}(\tau) =\left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4$$ In this form, \(j_{2A}\) is seen to be a 4th power. The function doesn't have a name, but is notable for the value \(j_{2A}(\tfrac12\sqrt{-58})=396^4\) in Ramanujan's well-known pi formula. Note that,
$$j_{4B}(\tau) = \sqrt{j_{2A}(2\tau)}\\j_{8C}(\tau) = \sqrt[4]{j_{2A}(2\tau)}$$
Level 3
$$\begin{align}
j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\
j_{3B}(\tau) &= \left(\frac{d_1}{d_3}\right)^{12}\\
j_{3C}(\tau) &= \sqrt[3]{j_{1A}(3\tau)}
\end{align}$$ Alternatively, $$j_{3A}(\tau) = \left(\frac{d_1^2}{d_3^2}+\frac{9\,d_9^3}{d_1\,d_3^2}\right)^6$$ In this form, \(j_{3A}\) is seen to be also a cube. An example is \(j_{3A}\big(\tfrac{1+\sqrt{-89/3}}{2}\big)=-300^3\) which can also be used in a Ramanujan-type pi formula. Notice that \(j_{3C}\) is a cube root.
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.
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.
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