TITLE: Modular parametrization of a curve of Heegner and Weber
QUESTION [5 upvotes]: The curve 
$$(X-16)^3=XY\tag{1}\label{1}$$
is essential to Heegner's approach to the class number one problem for imaginary quadratic fields. We have the following “modular” parametrization 
\begin{equation}\tag{2}\label{2}(X,Y)=\left(2^{12}\Phi(\tau),j(\tau)\right),\end{equation}
where $\Phi(\tau)=\frac{\Delta(2\tau)}{\Delta(\tau)}$. Note that the function $\Phi(\tau)$ is a Hauptmodul for the group $\Gamma_0(2)$.
My questions are:

The parametrization \eqref{2} can be deduced, quite laboriously, using Weierstrass elliptic functions and the product expansion for $\Delta$. However, when we clear the denominators, equation \eqref{1} (possibly) becomes an identity between modular forms and such identities should be easy to prove using the fact, that the space of modular forms has a finite dimension. Can this be done?
What can we say about $\Delta(2\tau)$? It is a modular form with respect to some group?

REPLY [2 votes]: We can also use Ramanujan's functions $Q(q), R(q) $ given by $$Q(q) =1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n},R(q)=1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\tag{1}$$ The expression $\Delta(q) $ is defined as $$\Delta(q) = Q^3(q)-R^2(q)\tag{2}$$  It is well known that $$\Delta(q) =1728\eta^{24}(q)\tag{3}$$ where by definition $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{4}$$ The invariant $j(q) $ is then defined by $$j(q) =\frac{1728 Q^{3}(q)}{Q^{3}(q)-R^{2}(q)}=\frac{Q^3(q)}{\eta^{24}(q)}\tag{5}$$ The relation we seek to verify can be written as $$(256\Delta(q^2)+\Delta(q))^3=\Delta(q^2)\Delta^2(q)j(q)$$ and replacing $q$ by $q^2$ we get $$ (256\Delta(q^4)+\Delta(q^2))^3=\Delta(q^4)\Delta^2(q^2)j(q^2)\tag{6}$$ Using identity $(3)$ and definition of $j(q) $ (via equation $(5)$) the above identity can be reduced to $$(256\eta^{24}(q^4)+\eta^{24}(q^2))^3=\eta^{24}(q^4)\eta^{24}(q^2)Q^{3}(q^2)$$ or $$256\eta^{24}(q^4)+\eta^{24}(q^2)=\eta^8(q^4)\eta^{8}(q^2)Q(q^2)\tag{7}$$ We can now use the link between these functions and elliptic integral with modulus $k$ corresponding to nome $q$. We have the following identities
$$\eta(q^2) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}k^{1/6}k'^{1/6}\tag{8a}$$
$$\eta(q^{4}) = 2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{8b}$$
$$Q(q^2) = \left(\frac{2K}{\pi}\right)^4(1-k^2+k^4)\tag{8c}$$ Using these identities we can see that both the LHS and RHS of equation $(7)$ can be written as $$2^{-8}\left(\frac{2K}{\pi}\right)^{12}k^4k'^2\left(1-k^2+k^4\right)$$ and the proof is thus complete.