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.

Comment #8537 by Mark on

should be for all $i\ge 0$ and $j>0$

Comment #9123 by on

No, this is correct as stated. For $i = 0$ one gets the cohomology of $\mathcal{F}$ which can be very interesting for any value of $j$!

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