$Pic$ of the stack of elliptic curves vs. $Pic$ of the coarse space

Question

There's a natural map $f:\overline{\mathcal{M}}_{1,1}\to \overline{M}_{1,1}\cong \mathbb{P}^1$ from the stack of elliptic curves to the coarse space. Both spaces have $Pic=\mathbb{Z}$ hence $f^*:\mathbb{Z}\to\mathbb{Z}$ is an homomorphism. What homomorphism? My guess is: $x \mapsto 24 x$ since the generator of the stack is the Hodge class, which has degree 1/24. Do you agree?

Answer 1

I believe the number is 12.

I will assume the characteristic of the base field is not 2 or 3 so that I can use $\overline{\mathcal{M}}_{1,1} \simeq \mathbf{P}(4,6)$. Recall that $\mathbf{P}(4,6)$ is constructed by dividing $V = \mathbf{A}^2 \smallsetminus \{ (0,0) \}$ by the weight $(4,6)$-action of $\mathbf{G}_m$.

The line bundle $\mathcal{O}(1)$ (which coincides with the Hodge bundle and generates the Picard group) can be constructed as the equivariant line bundle $V \times \mathbf{A}^1$ on $V$ with $\mathbf{G}_m$ acting by weights $(4,6,1)$. Then $\mathcal{O}(4)$ has a canonical section $g_4$ and $\mathcal{O}(6)$ has the section $g_6$. In $\mathcal{O}(12)$ we have the two sections

$1728 g_4^3$

$\Delta = g_4^3 - 27 g_6^2$.

This linear series defines $j : \mathbf{P}(4,6) \rightarrow \mathbf{P}^1$.

Note that $j$ has degree $1/2$ because of the generic automorphism on $\mathbf{P}(4,6)$. That is, $j_\ast j^\ast$ is multiplication by $1/2$. Thus the Hodge class $\lambda$ satisfies

$\int_{\overline{\mathcal{M}}_{1,1}} \lambda = \int_{\mathbf{P}(4,6)} c_1(\mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}(4,6)} c_1(j^\ast \mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}^1} j_\ast j^\ast c_1(\mathcal{O}(1)) = \frac{1}{24}$.

Another thing that may be confusing here is that $j$ is generically unramified, so that a local equation for a point in $\mathbf{P}^1$ pulls back under $j$ to a local equation in $\overline{\mathcal{M}}_{1,1}$. Thus $j^\ast \Delta = \delta$ (if $\Delta$ denotes the boundary in the coarse moduli space and $\delta$ the boundary in the stack).

Answer 2

Let us choose $\Delta_{irr}$ the point representing the class of a nodal curve as generator of $Pic(\overline{M}_{1,1})\cong\mathbb{Z}$. Let $\delta_{irr}$ be the corresponding boundary divisor in $\overline{\mathcal{M}}_{1,1}$, and let $f:\overline{\mathcal{M}}_{1,1}\rightarrow\overline{M}_{1,1}$ be the canonical morphism between the stack and its coarse moduli space.

Since a nodal curve $[C]\in\Delta_{irr}$ of arithmetic genus $1$ has two automorphisms (identity and elliptic involution) by Proposition $3.92$ of "Harris-Morrison Moduli of curves" we have: $$\delta_{irr}=\frac{1}{Aut(C)}f^{*}\Delta_{irr} = \frac{1}{2}f^{*}\Delta_{irr}.$$ Now, by Theorem 6.9 of Hain's notes:

http://arxiv.org/pdf/0812.1803v2.pdf

the Hodge class $\lambda$ generates $Pic(\overline{\mathcal{M}}_{1,1})$ and furthermore we have $$\mathcal{O}_{\overline{\mathcal{M}}_{1,1}}(\delta_{irr}) = 12\lambda\in Pic(\overline{\mathcal{M}}_{1,1}).$$ Finally $$f^{*}\Delta_{irr} = 2\delta_{irr} = 2\cdot12\lambda = 24\lambda$$ and the homomorphism $$f^{*}:Pic(\overline{M}_{1,1}) = \left\langle\Delta_{irr}\right\rangle\rightarrow Pic(\overline{\mathcal{M}}_{1,1}) = \left\langle\lambda\right\rangle$$ is given by $n\mapsto 24n$ as you predicted.

I guess that the underlying fact is that $\overline{\mathcal{M}}_{1,1}\cong\mathbb{P}(4,6)$. In order to pass to the coarse moduli space $\overline{M}_{1,1}\cong\mathbb{P}^{1}$ you have to take into account the two points of $\mathbb{P}(4,6)$ with stabilizers $\mathbb{Z}_{4}$ and $\mathbb{Z}_{6}$ ($lcm(4,6) = 12$) but also the fact that $\mathbb{P}(4,6)$ is not well-formed meaning that $4,6$ are both divided by $2$. So the general point of $\mathbb{P}(4,6)$ has stabilizer $\mathbb{Z}_{2}$ which corresponds to the elliptic involution of the general elliptic curve.