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:

  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?
  • $\begingroup$ Is there a typo in equation (1)? The curve given there is rational, hence not elliptic of course. $\endgroup$ – Peter Mueller May 19 '18 at 21:32
  • $\begingroup$ @Peter Mueller: There is no typo. It seems that I have forgotten what elliptic curves are. But the other two questions remain. $\endgroup$ – Shimrod May 19 '18 at 21:43

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 $\Gamma_0(2)$.

  • $\begingroup$ Yes, the curve should be $(X+16)^3=XY$. This approach seems simpler than using elliptic functions. $\endgroup$ – Shimrod May 20 '18 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.