# lifts of maps to $\mathcal{M}_{1,1}$

Hi,

here's there's a construction about elliptic curves that I do not completely understand. Suppose I consider the two following families of elliptic curves over $\mathbb{C}^*$.

The first, which I denote $F_0$ is the trivial family $E \times \mathbb{C}^*$, where $E$ is an elliptic curve with only $-Id$ as non-trivial automorphism. I will denote by $F_1$ the quotient of $F_0$ via the diagonal action of $-Id$, actiong both on $E$ and $\mathbb{C}^*$.

Remark that the quotient of the base is equal to $\mathbb{C}^*$ itself. Now $F_0$ has trivial monodromy around 0 and $F_1$ has not. The modular maps to $\mathcal{M}_{1,1}$ given by the two $F_i$ are constant, but there exist non-constant lifts depending on the description of the mod space.

The two lifts to the Siegel half space $\mathcal{H}_1$ (seeing the mod space of elliptic curves as a quotient by $SL_2(\mathbb{Z})$) are constant.

Let us now consider the quotient of $\mathbb{C}^2$ minus the discriminant (the Neil-parabola) by the torus $\mathbb{C}^*$ acting with weights $(4,6)$, i.e. let us parametrize the mod space with the coefficients $a$ and $b$ of the Weierstrass equation. The smooth curves are parmetrized by the open subspace $U\subset \mathbb{P}(4,6)$ complement of the point at infinity.

With this presentation, any lift of $F_0$ to $\mathbb{C}^2$ is the constant map to $(a,b)$, but a nonconstant lift of $F_1$ exists, namely and the non-constant map defined by

$$t \mapsto (t^2a, t^3 b).$$

Can you explain me why this happens? Is it related to the monodromy of the family? Or the connectedness of the group acting?

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Neither the Siegel upper half space nor the open subset of $\mathbb{P}(2,3)$ that you mention is the stack of elliptic curves. The stack of elliptic curves is, generically, a $\mathbb{Z}_2$-gerbe. –  Jason Starr Apr 16 at 14:28
generically - I agree - is a Z_2 - gerbe, but there are elliptic curves with bigger automorphism group, no? and I want to consider all smooth elliptic curves –  IMeasy Apr 16 at 15:28
ah sorry, I think I see what you mean. let me edit the question. –  IMeasy Apr 16 at 15:36