Is there an elliptic surface over $Y(1)$?

Question

Actually I have a few related questions.

Here, by $Y(1)$ I mean the affine $j$-line $\text{SL}_2(\mathbb{Z})\backslash\mathcal{H}$.

I know $Y(1)$ is only a coarse moduli space, so there isn't a universal elliptic curve over it, but does there exist an elliptic surface over it such that the fiber above every point is the elliptic curve corresponding to that point?

Now, I know that above the open set $Y(1) \setminus \{i, e^{2\pi i/3}\}$, there is such an elliptic surface. My second question is: Does there exist an open cover of $Y(1)$ such that above each open set in the cover, there exists an elliptic surface with the above property? (I'm not asking for the surfaces to glue.)

Complex analytically, one can consider the space: $$\mathcal{H}\times\mathbb{C}$$ On this space we have an action (on the right) by $\mathbb{Z}\times\mathbb{Z}$, acting via $$(\tau,x)\cdot(a,b) = (\tau, x + a\tau + b),$$ and an action (on the left) by $\text{SL}_2(\mathbb{Z})$, acting via $$\gamma\cdot(\tau,x) = (\gamma\tau, x)$$ where $\gamma$ acts on $\mathcal{H}$ by fractional linear transformations. Since intuitively, $\mathbb{Z}\times\mathbb{Z}$ acts on $\mathcal{H}\times\mathbb{C}$ "discretely", so the quotient $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ should be a complex manifold, and is essentially an elliptic surface over $\mathcal{H}$.

If $\Gamma \subset \text{SL}_2(\mathbb{Z})$, then we may also try to form the quotient $$\mathbb{E}(\Gamma) := \Gamma\backslash\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}.$$ Intuitively, if $\Gamma$ has no elliptic elements, then it should act "discretely" on $\mathcal{H}\times\mathbb{C}/\mathbb{Z}\times\mathbb{Z}$ and the quotient ought to be a manifold. Hence, if $\Gamma = \Gamma(2)$ (ie, matrices congruent to the identity mod 2), then since $\Gamma(2)$ has no elliptic elements, shouldn't $\mathbb{E}(\Gamma(2))$ be a (complex) manifold? If it is, then it's a complex manifold above $Y(2) := \Gamma(2)\backslash\mathcal{H}$, which is again only a coarse moduli scheme, and hence has no universal elliptic curve. In this case, my third question is: is $\mathbb{E}(\Gamma(2))$algebraic? and if it is, can someone describe heuristically how it's different from a universal elliptic curve over $Y(2)$, if one existed?

The context for these questions comes from me trying to understand why Katz's definition of modular forms for $\Gamma(N)$ (in his paper on $p$-adic modular forms) is properly a generalization of the analytic definition of modular forms. In particular, I'm trying to understand why they must give holomorphic functions on $\mathcal{H}$. I see why this must be the case when $N \ge 3$, since then you have a universal elliptic curve, and in this case holomorphicity is a result of the required compatibility with base change, but in the case of $N = 1$ and $N = 2$, I'm still a little confused. Relevant references would be appreciated as well.

thanks,

• will

Answer 1

1 The last exercise of Hartshorne chapter 4 section 4 proves that there is no elliptic surface over $\mathbb A^1$ with nonconstant $j$ invariant. Obviously this is a special case.

2 No as well. Put the fiber in Weirstrass form near $j=0$. Then one can write it as $y^2=x^3-g_2x-g_3$, where $g_2$ and $g_3$ are both functions of $j$ that are well-defined at $j=0$. Furthermore clearly $g_3$ is nonvanishing at $j=0$ and $g_2$ is vanishing at $j=0$. Then computing the $j$ invariant near $j=0$ as $g_2^3/(4g_2^3-27g_3^2)$ up to a constant, we see it vanishes to at least third order, an obvious falsehood.

3 Yes, you're thinking of the Legendre family $y^2=x(x-1)(x-\lambda)$. The Legendre family is universal up to a quadratic twist - every family with full level two structure is a pullback of Legendre, up to a quadratic twist. So it is as close an approximation to the universal elliptic curve over $Y(2)$ as you can get in the category of schemes.