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?

  • 1
    $\begingroup$ That seems reasonable. But how do you know that $Pic \mathcal M_{1,1} = \mathbb Z$? You have a proof or reference? $\endgroup$
    – Joël
    Commented Nov 1, 2013 at 17:06
  • 2
    $\begingroup$ If I remember correctly, Mumford only does it over a field not of characteristic 2 or 3. The general reference is Fulton and Olsson: arxiv.org/abs/0704.2214 $\endgroup$ Commented Nov 1, 2013 at 23:14
  • 2
    $\begingroup$ It is easy to see that the 12'th tensor power of the Hodge bundle--and no smaller tensor power--descends to the coarse moduli space: consider the possible automorphism groups of elliptic curves and their actions on the tangent space at the identity. $\endgroup$
    – naf
    Commented Nov 2, 2013 at 11:09
  • 2
    $\begingroup$ The general statement is, I think, the following: Let $f:\mathcal{X} \to X$ be the map from a DM-stack to its coarse moduli stack, $L$ a line bundle on $\mathcal{X}$ and $n$ the lowest common multiple of the orders of all automorphism groups. Then $f_*L^{\otimes n}$ is a line bundle and $L^{\otimes n} \to f^*f_*L^{\otimes n}$ an isomorphism. At least, I know a proof in the case that $\mathcal{X}$ is separated and of finite type over a noetherian base scheme. $\endgroup$ Commented Nov 4, 2013 at 15:17
  • 1
    $\begingroup$ I really would love to see a proof. Do you have a reference? $\endgroup$
    – IMeasy
    Commented Nov 4, 2013 at 20:34

2 Answers 2


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).

  • $\begingroup$ One can use specialisation to deduce the result over any field: both the Hodge bundle and $O_{\mathbb{P}^1}(1)$ are defined over $Spec(\mathbb{Z})$ and generate the relevant Picard groups $\endgroup$
    – naf
    Commented Nov 8, 2013 at 6:30

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:


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.

  • $\begingroup$ As pointed out by Jonathan Wise, your appeal to Proposition 3.92 of Harris-Morrison is not correct. $\endgroup$
    – naf
    Commented Nov 7, 2013 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.