Solving sextic polynomials with modular forms.

Question

A very long time ago, I made a since removed post about solving higher order polynomials. This post demonstrated the radical solutions for degrees $1$ to $4$ and also some solutions for higher degree polynomials with less familiar functions. It was intended to be far reaching while still being comprehensible by a beginner. However, there was $1$ formula I knew that didn't quite fit this bill. I tried to include it anyway, which led to the post being controversial and somewhat hated for involving higher functions at the end while still supposedly being beginner oriented. Therefore, I think it makes sense to make a separate post discussing this formula and how it might be generalized. $$x^{6}+10x^{3}+cx+5=0,\ x=\frac{5η\left(5j^{-1}\left(-c^{3}\right)\right)^{2}}{η\left(j^{-1}\left(-c^{3}\right)\right)^{2}}$$ This formula solves a special sextic using modular forms, and I started wondering. Could there be a general sextic formula in modular forms using a similar approach? If so, how so? I'm not entirely sure if I tagged this properly, kinda hard to tell since I haven't really seen anything like this before. I honestly don't even understand how I knew this back then. It's like it dropped out of the sky.

Answer 1

Consider the polynomial $$P(x)=^6+10^3+(-j)^{\frac13}+5$$ Then $P(x)=0$ can be solved for $j$ as $$j=\frac{\left(x^6+10 x^3+5\right)^3}{x^3}.$$ We can now lift both $j$ and $x$ to be functions of $\tau\in\mathbb H$, with $j$ the modular $j$-function. We can expand $j$ around $x=0$ and invert the series order by order. Inserting the $q=e^{2\pi i\tau}$-expansion of $j$, one finds $$x(\tau)=5q^{\frac13}(1+2 q+5 q^2+10 q^3+20 q^4+...),$$ which agrees with the Fourier expansion of the inverse of the McKay-Thompson series of class 15D for the Monster group. Thus $$x(\tau)=5\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^2.$$ By defining $c=(-j(\tau_c))^{\frac13}$, $P(x(\tau_c))=0$ has then one root $x(\tau_c)=5\left(\frac{\eta(5\tau_c)}{\eta(\tau_c)}\right)^2$.