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.

  • $\begingroup$ You have to be careful here. I am not sure that you have written down a family of elliptic curves in general. For example if $-1\in\Gamma$ then you seem to be identifying $(\tau,x)$ with $(\tau,-x)$. There may be similar problems with elliptic points even if $-1\not\in\Gamma$ (for example if $\Gamma=\Gamma_1(3)$). The problem is that your moduli problem is not representable in general, and the way you've set it up I think that in the curve above $\tau$ is the torus you think it is, modulo e.g. its automorphism group, which from what you write is I think not what you want to happen. $\endgroup$ – Kevin Buzzard May 24 '12 at 18:22
  • $\begingroup$ I should of course say that my comments do not pertain to the question (which restricts to the torsion-free case); they just pertain to the background stuff. $\endgroup$ – Kevin Buzzard May 24 '12 at 18:24
  • $\begingroup$ You are right I was a bit careless in the set up $\endgroup$ – Hugo Chapdelaine May 24 '12 at 19:34
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    $\begingroup$ In general, if $X$ is an algebraic variety over $\mathbb{C}$ and $E\to X$ is an analytic family of elliptic curves, then $E$ is automatically algebraic, and in fact projective over $X$. In particular, if $X$ is quasi-projective, then so is $E$. The point is that the Weierstrass embedding of an elliptic curve into $\mathbb{P}^2$ works in families as well. $\endgroup$ – Keerthi Madapusi Pera May 26 '12 at 16:22
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    $\begingroup$ Keerthi, on the contrary, the Hopf surface is the easiest example of a non-algebraic surface. It is an analytic surface $(\mathbb C^2-0) / q^{\mathbb Z}$ with evident map to $P^1=(\mathbb C^2-0)/\mathbb C^*$ exhibiting it as a bundle with fiber all elliptic curves isomorphic to $\mathbb C^\* / q^{\mathbb Z}$, but its function field is one dimensional. en.wikipedia.org/wiki/Hopf_surface $\endgroup$ – Ben Wieland May 27 '12 at 21:49

Here is an alternative take on Donu's argument: Removing the image $\sigma$ of a section of $E_\Gamma$ allows one to regard a fiber of $E_{\Gamma}$ as a once-punctured torus $S$. (In order to construct a section, use the standard upper half-space model of the Teichmuller space of the tori, so that each marked torus is identified with the fundamental parallelogram $P$ with vertices $0, 1, \omega, \omega+1\in {\mathbb C}$. Then the section is given by the map $\omega\to 0\in P$.) Then $M=E_{\Gamma}\setminus \sigma$ is isomorphic to the quotient of the Teichmuller space $T(S)$ of $S$ by a finite-index torsion-free subgroup $Mod^o_S$ of the mapping class group $Mod_S$ of $S$. Now, this is a general fact (Deligne-Mumford, et al) that $T(S)/Mod^o_S$ is quasi-projective (for any Riemann surface of finite type). It is not hard to see that DM compactification of $T(S)/Mod^o_S$ is our case will add (among other things) the curve $\sigma$ back to $M$, thus, providing a projective compactification of $M$.

In the special case you are interested in, it seems that quasi-projectivity of $E_{\Gamma}$ was first proven by Kodaira (On compact analytic surfaces. II, III), at least, Shioda (On elliptic modular surfaces, 1972) attributes the result to him.

Addendum: As an alternative to this argument, one can use Ron Livne's thesis "On certain covers of the universal elliptic curve". Livne proves that for every level $N\ge 5$ congruence subgroup $\Gamma(N)$ in the modular group $SL(2, {\mathbb Z})$, the universal elliptic curve over $E(N):={\mathbb H}^2/\Gamma(N)$ admits a degree $d\ge 2$ cyclic branched cover $E_d(N)$, so that the latter admits a compactification $X_d(N)$ (compatible with branched cover), so that $X_d(N)$ is a general type projective surface. Thus, every $E_\Gamma$ admits a finite regular cover which is biholomorphic to a finite cover over one of the $E_d(N)$'s. Thus, $E_\Gamma$ is quasi-projective.

Livne also refers to Mumford's paper "Prym varieties. I." Contributions to analysis (a collection of papers dedicated to Lipman Bers), pp. 325–350. Academic Press, New York, 1974, for a direct proof of quasi-projectivity of $E(N)$'s. I do not have access to Mumford's paper, so I cannot say for sure.

  • $\begingroup$ @Misha, do you have a nice reference (which covers the case that I'm interested in) which shows in details why is $T(S)/Mod_S^{\circ}$ algebraic. I would like to see how one uses torsion-freeness in order to deduce algebraicity, which is the whole point in my question. $\endgroup$ – Hugo Chapdelaine May 29 '12 at 20:04
  • $\begingroup$ One can use S.Wolpert "On obtaining a positive line bundle from the Weil-Petersson class," Amer. J. Math. 107 (1985), no. 6, 1485–1507 (1986). In this paper Wolpert proves that the DM compactification of the moduli space is projective by a local computation, proving positivity of current defined by Weil-Peterson metric, and, hence, positivity of the corresponding line bundle. Since the computation is local, it goes through for finite-index subgroups of the mapping class group. I am also adding an alternative reference for the main result in my main answer. $\endgroup$ – Misha May 29 '12 at 21:42
  • $\begingroup$ @Hugo: As an alternative to differential-geometric arguments you can use the following: The DM compactification $\bar{M}_{g,n}$ of the moduli space $M_{g,n}=T_{g,n}/Mod_{g,n}$ is projective. Hence, it admits a positive line bundle $L$ (here I regard $\bar{M}_{g,n}$ as a complex space rather than a stack). For every finite-index subgroup $Mod^o_{g,n}\subset Mod_{g,n}$, the quotient $M^o_{g,n}=T_{g,n}/Mod^o_{g,n}$ admits a compactification to a complex space $\bar{M}^o_{g,n}$ and a holomorphic branched cover $f: \bar{M}^o_{g,n}\to \bar{M}_{g,n}$. ... $\endgroup$ – Misha May 30 '12 at 4:25
  • $\begingroup$ Pull-back bundle $f^*(L)$ on $\bar{M}^o_{g,n}$ is still positive, so $\bar{M}^o_{g,n}$ is projective (by Kodaira) provided that it is a complex manifold. The latter you get by taking $Mod^o_{g,n}$ to be any finite-index subgroup in an appropriate congruence-subgroup in $Mod_{g,n}$ (as in Looijenga's paper on Prym level structures staff.science.uu.nl/~looij101/prymlevel2.ps). Projectivity of the rest of quotients of $T_{g,n}$ by finite-index subgroups in $Mod_{g,n}$ follows by taking finite quotients of $M^o_{g,n}$'s above. $\endgroup$ – Misha May 30 '12 at 4:42
  • $\begingroup$ @Misha, so what you told me in the comments look good, I like this idea of using an explicit positive line bundle. So concerning your addendum keep in mind that not all finite index subgroups are congruence subgroups. So I don't quite understand why your $X_d(N)$ is projective... $\endgroup$ – Hugo Chapdelaine May 30 '12 at 10:38

The point is that by your construction $f:\mathcal{E}_\Gamma\to Y_\Gamma$ is an elliptic surface with a section $\sigma$. So as Keerthi said in the comments, $f$ is projective. In fact the relative divisor $3\sigma$ gives the Weirstrass embedding $\mathcal{E}_\Gamma\to \mathbb{P}(f_*\mathcal{O}(3\sigma))$). (Ben's comment shows that projectivity can fail without a section.) From here, quasiprojectivity is straight forward: If $V$ is an extension of $f_*\mathcal{O}(3\sigma)$ to a vector bundle on the smooth projective closure of $ Y_\Gamma$, then the closure of $\mathcal{E}_\Gamma$ in $\mathbb{P}(V)$ is a projective variety. Therefore $\mathcal{E}_\Gamma$ is quasiprojective.

Given the endless stream of comments, perhaps I should add a few words of clarification:

  1. Since $Y=Y_\Gamma$ is noncompact, any vector bundle such as $f_*\mathcal{O}(3\sigma)$ is in fact (analytically) trivial. This is definitely overkill, but you can use Grauert, "Analytichse Faserungen..." Math. Ann 1958.
  2. Thus $\mathbb{P}(\mathcal{O}(3\sigma))\cong Y\times \mathbb{P}^2$. This embeds into $\bar{Y}\times \mathbb{P}^2$, where $\bar Y$ is a (the) smooth projective compactification of $Y$.
  3. We can take the closure of $\mathcal{E}_\Gamma$ to get a projective variety, and the quasiprojectivity of this family follows easily.
  4. Note that the fibres of the closure $\overline{\mathcal{E}_\Gamma}$ may be singular.
  5. If the Hopf surface had a section, it would lift to a rational curve in $\mathbb{C}^2-\lbrace 0\rbrace$. Well, I'll let you think about why that might be a problem.

OK, I guess there some issues.... I'm converting this to CW. Anyone, who wants to fix this is welcome to. I've got to finishing my refereeing....

  • $\begingroup$ Donu, I think I'm missing something important here. Say that you take the universal elliptic curve $E\rightarrow \mathbf{C}-\{0,1728\}$, then you cannot extend your vector bundle over the smooth projective closure of $\mathbf{C}-\{0,1728\}$ which is $\mathbf{C}$, since there is no universal elliptic curve over $\mathbf{C}$. So may be you could develop a little bit what you mean when you say extending the vector bundle to the smooth projective locus. $\endgroup$ – Hugo Chapdelaine May 29 '12 at 19:10
  • $\begingroup$ the 3rd $\mathbf{C}$ should be replaced by $\mathbb{P}^1(\mathbf{C})$ of course. $\endgroup$ – Hugo Chapdelaine May 29 '12 at 20:02
  • $\begingroup$ Hugo, these are separate issues. A vector bundle on a punctured disk is trivial, so it extends. So applying this remark and the above construction, we see $E$ extends to a family of possibly singular cubics over $0$ and $1728$. $\endgroup$ – Donu Arapura May 29 '12 at 20:05
  • $\begingroup$ Well holomorphic vector bundles over (connected) non-compact Riemann surfaces are trivial in the analytic category since the construction of sections which trivialize your bundle is transcendental but I don't see why it should trivialize in the algebraic category. $\endgroup$ – Hugo Chapdelaine May 29 '12 at 21:20
  • $\begingroup$ @Donu, I'm a bit puzzle by what you said about the existence of a section. Take $Y=(\mathbf{C}^2-(0,0))/<(2,2)>$, then it seems to me that the map $s:\mathbf{P}^1\rightarrow Y$ given by $s(x)=(2,2)$ gives you section of your elliptic fibration. In this case, all fibers are isomorphic to the elliptic curve $\mathbf{C}^{\times}/<2>$. $\endgroup$ – Hugo Chapdelaine May 29 '12 at 21:23

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