Modular parametrization of a curve of Heegner and Weber

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 $$\tag{2}\label{2}(X,Y)=\left(2^{12}\Phi(\tau),j(\tau)\right),$$ 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:

1. 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?
2. What can we say about $\Delta(2\tau)$? It is a modular form with respect to some group?
• Is there a typo in equation (1)? The curve given there is rational, hence not elliptic of course. – Peter Mueller May 19 '18 at 21:32
• @Peter Mueller: There is no typo. It seems that I have forgotten what elliptic curves are. But the other two questions remain. – Shimrod May 19 '18 at 21:43
• Such identities can also be proved by using relation between these forms and elliptic integrals and moduli. The algebra involved should not be more than than needed in evaluation of coefficients in the modular forms. – Paramanand Singh Jul 5 at 7:30

Firstly, you have a typo. The left side is $$(X-16)^3=-4096 + 3145728q - 729808896q^2+O(q^3)$$ while the right side is $$XY=4096 + 3145728q + 880803840q^2+O(q^3).$$
Looking at (19) of Stark's paper, I think the relevant root of $(X-16)^3=Xj(\tau)$ is $X=-f_2(\tau)^{24}$, which is $-2^{12}\Phi(\tau)$ in your notation.
Proceeding from this, you are then interested in a modular forms proof of $$(2^{12}\Delta(q^2)+16\Delta(q))^3=2^{12}\Delta(q^2)\Delta(q)^2j(q).$$ I'm not sure there is anything better than noting that both sides are weight 36 modular forms of level 2, and then equating enough coefficients to exploit the finite dimensionality. Whether this is "easier" than the method you mention is not clear.
I don't know what you intend by (2). In general, $f(\tau)$ is a modular form on $\Gamma_0(N)$, then $f(M\tau)$ is a modular form on $\Gamma_0(MN)$. So clearly $\Delta(2\tau)$ is a modular form on $\Gamma_0(2)$. Indeed, as you note, the quotient is a modular function on $\Gamma_0(2)$.
• Yes, the curve should be $(X+16)^3=XY$. This approach seems simpler than using elliptic functions. – Shimrod May 20 '18 at 17:59
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.