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# On the algberaicityofthe 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 $$\mathbf{C}/(\mathbf{Z}+\tau\mathbf{Z})\rightarrow \mathbf{C}/((a\tau+b)\mathbf{Z}+(c\tau+d)) \mathbf{Z}\rightarrow (\mathbf{C}/\mathbf{Z}+\gamma\tau\mathbf{Z}) \;\;\;\; (*)$$ where the first map is the identity and the second map is the multiplication by $(c\tau+d)^{-1}$. Let $$\tilde{\mathcal{E}}_{\Gamma}=\{(\tau,x):\tau\in\mathbb{H},x\in \mathbf{C}/(\mathbf{Z}+\tau\mathbf{Z})\}$$ We have a natural left action of $\Gamma$ on $\tilde{\mathcal{E}}_{\Gamma}$ given by $$\gamma(\tau,x)=(\gamma\tau,j(\gamma,\tau)^{-1}x),$$ 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): $$\pi_\Gamma:\Gamma\backslash\tilde{\mathcal{E}}_{\Gamma}=:\mathcal{E}_{\Gamma}\rightarrow Y_{\Gamma}:=\Gamma\backslash \mathbb{H}$$ 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 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 $\mathcal{E}_{\Gamma}=(\mathcal{E}_{\Gamma'})^{\Gamma/\Gamma'}$ we readily see that if $\mathcal{E}_{\Gamma'}$ is affine then automatically $\mathcal{E}_{\Gamma}$ is affine being the quotient of affine variety by a finite group.

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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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