Lemma 60.21.2. Assumptions and notation as in Proposition 60.21.1. Then
\[ H^ j(\text{Cris}(X/S), \mathcal{F} \otimes _{\mathcal{O}_{X/S}} \Omega ^ i_{X/S}) = 0 \]
for all $i > 0$ and all $j \geq 0$.
Proof. Using Lemma 60.12.6 it follows that $\mathcal{H} = \mathcal{F} \otimes _{\mathcal{O}_{X/S}} \Omega ^ i_{X/S}$ also satisfies assumptions (1) and (2) of Proposition 60.21.1. Write $M(n)_ e = \Gamma ((X, T(n)_ e, \delta (n)), \mathcal{F})$ so that $M(n) = \mathop{\mathrm{lim}}\nolimits _ e M(n)_ e$. Then
\begin{align*} \mathop{\mathrm{lim}}\nolimits _ e \Gamma ((X, T(n)_ e, \delta (n)), \mathcal{H}) & = \mathop{\mathrm{lim}}\nolimits _ e M(n)_ e \otimes _{D(n)_ e} \Omega _{D(n)}/p^ e\Omega _{D(n)} \\ & = \mathop{\mathrm{lim}}\nolimits _ e M(n)_ e \otimes _{D(n)} \Omega _{D(n)} \end{align*}
By Lemma 60.19.3 the cosimplicial modules
\[ M(0)_ e \otimes _{D(0)} \Omega ^ i_{D(0)} \to M(1)_ e \otimes _{D(1)} \Omega ^ i_{D(1)} \to M(2)_ e \otimes _{D(2)} \Omega ^ i_{D(2)} \to \ldots \]
are homotopic to zero. Because the transition maps $M(n)_{e + 1} \to M(n)_ e$ are surjective, we see that the inverse limit of the associated complexes are acyclic1. Hence the vanishing of cohomology of $\mathcal{H}$ by Proposition 60.21.1. $\square$
[1] Actually, they are even homotopic to zero as the homotopies fit together, but we don't need this. The reason for this roundabout argument is that the limit $\mathop{\mathrm{lim}}\nolimits _ e M(n)_ e \otimes _{D(n)} \Omega ^ i_{D(n)}$ isn't the $p$-adic completion of $M(n) \otimes _{D(n)} \Omega ^ i_{D(n)}$ as with the assumptions of the lemma we don't know that $M(n)_ e = M(n)_{e + 1}/p^ eM(n)_{e + 1}$. If $\mathcal{F}$ is a crystal then this does hold.
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8537 by Mark on
Comment #9123 by Stacks project on