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, $$j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}$$