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On the algberaicity of the universal elliptic curve associated to a torsion free subgroup |
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So let $\Gamma\subseteq SL_2(\mathbf{Z})$ be a finite index subgroup (not necessarily a congruence subgroup). Recall that we have an action of $SL_2(\mathbf{R})$ on
$\mathbb{H}=\{z\in\mathbf{C}:\Im(z)>0\}$ by moebius transformations and therefore of $\Gamma$. If $\tau\in\mathbb{H}$ and $[a,b,c,d]=\gamma\in SL_2(\mathbf{Z})$ then we have an isomorphism
of complex tori
In general one always has that $Y_{\Gamma}$ is a quasi-projective curve defined over $\mathbf{C}$ (in fact it is always possible to define this curve over $\overline{\mathbf{Q}}$, the algebraic closure of $\mathbf{Q}$). [For example, when $\Gamma=\Gamma_0(N)$ one may look at the modular polynomial $f_N(x,y)\in\mathbf{Z}[x,y]$. Then using the modular interpretation one can show that there exists a (complex analytic) immersion of $Y_{\Gamma_0(N)}$ onto the plane curve $C_N: f_N(x,y)=0$. If I remember correctly, in the finite chart $\mathbf{C}^2$ (for $N$ large enough) the singularities of $C_N$ are nodes and thus one can blow-up these points (over $\mathbf{Q}$!). From this one can construct a complex analytic isomorphism between $Y_{\Gamma_0(N)}$ and the blow-up (which is quasi-projective curve defined over $\mathbf{Q}$.] So here are 2 natural questions. Q1: Is $\mathcal{E}_{\Gamma}$ quasi-projective (at least when $\Gamma$ is torsion free)? Q2: If the answer is yes, then what is the cleanest (and if possible most transparent) way of showing that $\mathcal{E}_{\Gamma}$ is quasi-projective? So for the second question, the thing that I have in mind would be to 1) construct some complex analytic immersion $\pi:\mathcal{E}_{\Gamma}\rightarrow Z$, where $Z$ is a quasi-projective surface and 2) performing a sequence of blow-ups on $Z$ I would try to construct an embedding. added: Note that one can always find a normal finite index subgroup $\Gamma'\leq \Gamma$.
Since |
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which is just a reinterpretation of $(*)$. Here $j(\gamma,\tau)=c\tau+d$. We thus get the following family of curves (note that the fibers are not necessarily elliptic curves because of the presence of torsion in $\Gamma$ as K. Buzzard pointed out):In the case where $\Gamma$ is torsion free, we readily see that the fibers are elliptic curves and the the $\Gamma$ action is compatible with the addition on the $tori$ (this somehow justifies the terminology "universal elliptic curve" over $Y_{\Gamma}$).
In general one always have has that $Y_{\Gamma}$ is a quasi-projective curve defined over $\mathbf{C}$ (in fact it is always possible to define this curve over $\overline{\mathbf{Q}}$). \overline{\mathbf{Q}}$, the algebraic closure of $\mathbf{Q}$). [For example, when $\Gamma=\Gamma_0(N)$ one may look at the modular polynomial $f_N(x,y)\in\mathbf{Z}[x,y]$. Then using the modular interpretation one can show that there exists a (complex analytic) immersion of $Y_{\Gamma_0(N)}$ onto the plane curve $C_N: f_N(x,y)=0$. If I remember correctly, in the finite chart $\mathbf{C}^2$ (for $N$ large enough) the singularities of $C_N$ are nodes and thus one can blow-up these points (over $\mathbf{Q}$!). From this one can construct a complex analytic isomorphism between $Y_{\Gamma_0(N)}$ and the blow-up (which is quasi-projective scheme curve defined over $\mathbf{Q}$.] In general, even though $Y_{\Gamma}$ is quasi-projective one does not necessarily have that $\mathcal{E}_{\Gamma}$ is a quasi-projective surface. For instance this happens when one takes $\Gamma=SL_2(\mathbf{Z})$. [ here is a rough sketch on how to show that $\mathcal{E}_{SL_2(\mathbf{Z})}$ is not quasi-projective: Let $A=\mathbf{C}[\frac{1}{\lambda},\frac{1}{1-\lambda}]$ be the ring of regular functions of $A_{\mathbf{C}}^1=Y_{SL_2(\mathbf{Z})}-\{0,1\}$. Then there is an action of $S_3$ on $A$ given by $\lambda\mapsto\frac{1}{\lambda}$ and $\lambda\mapsto\frac{1}{1-\lambda}$. Now let $E$ be the generic fiber of $\mathcal{E}_{SL_2(\mathbf{Z})}\rightarrow Spec(A)$ and show that this action on $A$ lift to an action on $E$. Finally doing a local computation on $E[2]$ show that this implies that $\sqrt{\lambda}\in A$ which is absurd.] Q1: If Is $\mathcal{E}_{\Gamma}$ quasi-projective (at least when $\Gamma$ is torsion free, is $\mathcal{E}_{\Gamma}$ quasi-projectivefree)? |
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