Levels 14 & 28

For certain even levels divisible by \(7\), the situation is now different. We can still find an eta quotient \(a\) such that \(a+m = b\) is also an eta quotient, but only for one rational \(m\).

I. Moonshine functions: Define $$a(\tau) =\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right),\quad b(\tau) = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$ then \(a+2 = b\) or, $$\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right)+2 = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$ for the unique \(m=2\). This is the single trinomial identity of level 28. First, we have the 3 moonshine functions of level 14,
$$\begin{align}
j_{14A}(\tau) &= \left(\sqrt{j_{14C}}+\frac{1}{\sqrt{j_{14C}}}\right)^2\\
j_{14B}(\tau) &= \left(\frac{d_1\,d_7}{d_2\,d_{14}}\right)^3\\
&= b+\frac{4}{b}-4\\
j_{14C}(\tau) &= \left(\frac{d_2\,d_7}{d_1\,d_{14}}\right)^4
\end{align}$$ and the 4 moonshine functions of level 28,
$$\begin{align}\quad
j_{28A}(\tau) &= \sqrt{j_{14A}(2\tau)} = j_{28D} (\tau)+\frac{1}{j_{28D} (\tau)}\\
j_{28B}(\tau) &= \left(\frac{d_2^2\,d_{14}^2}{d_1\,d_4\,d_7\,d_{28}}\right)^3\\
&= a+\frac{4}{a}+4\\
\color{red}{j_{28C}}(\tau) &=  \frac{d_1\,d_7}{d_4\,d_{28}}\;=\; a\\
j_{28D}(\tau) &=  \sqrt{j_{14C}(2\tau)} = \left(\frac{d_4\,d_{14}}{d_2\,d_{28}}\right)^2
\end{align}$$ Note that \(j_{14B}\big(\tfrac12+\tau\big) = -j_{28B}(\tau)\).