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:
Idea. Local properties of an algebraic stack are encoded in the local properties of its smooth cover.
Global properties:
Idea. Some global properties of an algebraic stack can be read off from the corresponding property of a suitable1 smooth cover.
Quasi-coherent sheaves:
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:
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
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
Idea. Note that
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.
Comments (4)
Comment #9009 by James on
Comment #9010 by James on
Comment #9015 by James on
Comment #9196 by Stacks project on