## 105.7 Properties of algebraic stacks

Ok, so now we know that $\mathcal{M}_{1, 1}$ is an algebraic stack. What can we do with this? Well, it isn't so much the fact that it is an algebraic stack that helps us here, but more the point of view that properties of $\mathcal{M}_{1, 1}$ should be encoded in the properties of morphisms $S \to \mathcal{M}_{1, 1}$, i.e., in families of elliptic curves. We list some examples

Local properties:

$\mathcal{M}_{1, 1} \to \mathop{\mathrm{Spec}}(\mathbf{Z})\text{ is smooth} \Leftrightarrow W \to \mathop{\mathrm{Spec}}(\mathbf{Z})\text{ is smooth}$

Idea. Local properties of an algebraic stack are encoded in the local properties of its smooth cover.

Global properties:

$\begin{matrix} \mathcal{M}_{1, 1}\text{ is quasi-compact} \Leftarrow W\text{ is quasi-compact} \\ \mathcal{M}_{1, 1}\text{ is irreducible} \Leftarrow W\text{ is irreducible} \end{matrix}$

Idea. Some global properties of an algebraic stack can be read off from the corresponding property of a suitable1 smooth cover.

Quasi-coherent sheaves:

$\mathit{QCoh}(\mathcal{O}_{\mathcal{M}_{1, 1}}) = H\text{-equivariant quasi-coherent modules on }W$

Idea. On the one hand a quasi-coherent module on $\mathcal{M}_{1, 1}$ should correspond to a quasi-coherent sheaf $\mathcal{F}_{S, e}$ on $S$ for each morphism $e : S \to \mathcal{M}_{1, 1}$. In particular for the morphism $(E_ W, f_ W, 0_ W) : W \to \mathcal{M}_{1, 1}$. Since this morphism is $H$-equivariant we see the quasi-coherent module $\mathcal{F}_ W$ we obtain is $H$-equivariant. Conversely, given an $H$-equivariant module we can recover the sheaves $\mathcal{F}_{S, e}$ by descent theory starting with the observation that $S \times _{e, \mathcal{M}_{1, 1}} W$ is an $H$-torsor.

Picard group:

$\mathop{\mathrm{Pic}}\nolimits (\mathcal{M}_{1, 1}) = \mathop{\mathrm{Pic}}\nolimits _ H(W) = \mathbf{Z}/12\mathbf{Z}$

Idea. We have seen the first equality above. Note that $\mathop{\mathrm{Pic}}\nolimits (W) = 0$ because the ring $\mathbf{Z}[a_1, a_2, a_3, a_4, a_6, 1/\Delta ]$ has trivial class group. There is an exact sequence

$\mathbf{Z}\Delta \to \mathop{\mathrm{Pic}}\nolimits _ H(\mathbf{A}^5_{\mathbf{Z}}) \to \mathop{\mathrm{Pic}}\nolimits _ H(W) \to 0$

The middle group equals $\mathop{\mathrm{Hom}}\nolimits (H, \mathbf{G}_ m) = \mathbf{Z}$. The image $\Delta$ is $12$ because $\Delta$ has degree $12$. This argument is roughly correct, see [PicM11].

Étale cohomology: Let $\Lambda$ be a ring. There is a first quadrant spectral sequence converging to $H^{p + q}_{\acute{e}tale}(\mathcal{M}_{1, 1}, \Lambda )$ with $E_2$-page

$E_2^{p, q} = H_{\acute{e}tale}^ q(W \times H \times \ldots \times H, \Lambda ) \quad (p\text{ factors }H)$

Idea. Note that

$W \times _{\mathcal{M}_{1, 1}} W \times _{\mathcal{M}_{1, 1}} \ldots \times _{\mathcal{M}_{1, 1}} W = W \times H \times \ldots \times H$

because $W \to \mathcal{M}_{1, 1}$ is a $H$-torsor. The spectral sequence is the Čech-to-cohomology spectral sequence for the smooth cover $\{ W \to \mathcal{M}_{1, 1}\}$. For example we see that $H^0_{\acute{e}tale}(\mathcal{M}_{1, 1}, \Lambda ) = \Lambda$ because $W$ is connected, and $H^1_{\acute{e}tale}(\mathcal{M}_{1, 1}, \Lambda ) = 0$ because $H^1_{\acute{e}tale}(W, \Lambda ) = 0$ (of course this requires a proof). Of course, the smooth covering $W \to \mathcal{M}_{1, 1}$ may not be “optimal” for the computation of étale cohomology.

[1] I suppose that it is possible an irreducible algebraic stack exists which doesn't have an irreducible smooth cover – but if so it is going to be quite nasty!

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