# Finding a pencil of elliptic curves parametrized by a given modular surface

The following is an attempt to formulate a couple of questions which have been lurking in the back of my mind for a while. I'm sorry if this is long, or if my terminology is not correct, or if my notation is non-standard, or if it turns out that I didn't understand a single thing that I thought I understood.

Suppose we are given a genus $0$ congruence subgroup $G$ of $\Gamma=PSL_2(\mathbb{Z})$.

The (compactified) modular surface $X_G={\mathbb{H}/G}$ is then a degree $n=|\Gamma:G|$ covering of $X=\mathbb{H}/\Gamma$. Let $K'=\mathbb{C}(w)$ denote the function field of $X_G$ and $K=\mathbb{C}(j)$ denote the function field of $X$. If $G$ is normal in $\Gamma$, the field extension $K'/K$ is a Galois extension with Galois group $\Gamma/G$. Klein's $j$ function can be written as a rational function in $w$, of degree $n$.

Now each elliptic curve $E/\mathbb{C}$ corresponds to a point of $X$. We can write such a curve explicitly as the locus of zeroes of a cubic in $\mathbb{CP}^2$; its coefficients are rational in $j$. For almost all choices of $E$, there are $n$ points of $X_G$ lying over $E$. In some cases, these points are known to correspond to some pairs $(E, S)$, where $S$ is some sort of substructure of $E$ (for example a "level structure").

What's not completely obvious to me is that such pairs $(E,S)$ should also be given by a cubic or quadric equation in normal form, whose coefficients are rational in $w$. Here's an example. The group $\Gamma(2)$ is normal in $\Gamma$; the quotient $\Gamma/\Gamma(2)$ is isomorphic to $S_3$. The projection map $X_{\Gamma(2)} \to X$ is a six-sheeted ramified covering. The generator for $K'$ is then the modular function $\lambda$. The points of $X_{\Gamma(2)}$ correspond to pairs $(E, S)$, where $S$ is a basis (over $\mathbb{F}_2$) for the $2$-torsion subgroup of $E$. Since this group is isomorphic to the Klein four-group, there are indeed $6$ different choices of bases for it over $\mathbb{F}_2$.

Now the fascinating thing is that we can consider such a curve as given by an equation in Legendre normal form $y^2=z(z-1)(z-\lambda)$. For most choices of $\lambda$, there are precisely six curves, in this form, which are isomorphic to this one; the other corresponding values of $\lambda$ are related by the action of $\Gamma/\Gamma(2)$.

Now my first question is: how would one "discover" the normal form $y^2=z(z-1)(z-\lambda)$, given only the group $\Gamma(2)$? Assume, perhaps, that we know (at least) the normal form associated to $j$. Given any subgroup $G$ of $\Gamma$, can we find a normal form associated to $G$ in this manner? If the genus of $X_G$ is bigger than $0$, can we hope, perhaps, to find a similar construction?

My second question is a bit more vague: what are other modular surfaces moduli spaces for? This is understood in some specific cases, but is a general answer to be hoped for? Are the examples which are understood well understood? Can we hope to associate to each modular surface a normal form, in which the extra "structure" $S$ (whatever it may be) is encoded?

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$@$Bruno: This is possibly a question for MathOverflow. (I can say something about it, although maybe not right now. There are a lot of other professionals on that site who would have useful things to say.) –  Pete L. Clark Jul 8 '11 at 2:31
Thank you Pete. I'll leave it on here for a few days and then post it there if I get no answer. –  Bruno Joyal Jul 9 '11 at 19:44
I see that Quiaochu Yuan asked pretty much exactly the same question on MathOverflow last year, and the answer is given! mathoverflow.net/questions/17031/… –  Bruno Joyal Jul 23 '11 at 6:48

Regarding how one would discover the Legendre form: As far as a moduli problem goes, $X(2)$ is supposed to capture elliptic curves $E$ together with a choice of basis for the $2$-torsion of $E$. If $y^2 = f(x)$ is an elliptic curve where $f(x)$ is a cubic, then the $2$-torsion points are given by $[x,y] = [e_i,0]$ where $e_i$ are the roots of $f(x)$. Now let's try to write down a parametrization of all such pairs. We can start by writing $y^2 = (x-e_1)(x-e_2)(x-e_3)$, and consider the parameter space $\mathbf{A}^3 \setminus \Delta$ given by the $e_i$, where we have removed the discriminant locus $\Delta$ given by $(e_1 - e_2)(e_1 -e_3)(e_2 - e_3) = 0$. Here we have our elliptic curve, together with our fixed basis of $2$-torsion generated (say) by $e_1$ and $e_2$. Yet clearly we can make a linear change of basis to the RHS to force $e_1 = 0$ and $e_2 = 1$. Then we get $y^2 = x(x-1)(x - \lambda)$ with $$\lambda = \frac{e_3 - e_1}{e_2 - e_1}$$ and $\lambda \neq 0,1$. This is basically it! From this description, we see that $\lambda$ gives us a pair $(E,E[2] \simeq (\mathbf{Z}/2)^2)$. We also see that the action of $\Gamma/\Gamma(2) = \mathrm{GL}_2(\mathbf{F}_2) = S_3$ acts on $X(2)$ just by permuting the $e_i$, which we can detect as acting on $\lambda$ in the usual ways $\lambda \mapsto 1/\lambda$, $\lambda \mapsto 1/(1 - \lambda)$, etc.
Thanks for your answer! However, you seem to be assuming that we know already what $X(2)$ is the moduli space for. My question is more along the lines of : given only the group $\Gamma(2)$, how might one discover this interpretation for $X(2)$ (along with the Legendre normal form), without any prior knowledge of what $X(2)$ should be? –  Bruno Joyal Jul 9 '11 at 19:44