Lemma 60.15.1. In Situation 60.7.5. Let $\mathcal{F}$ be a crystal in $\mathcal{O}_{X/S}$-modules on $\text{Cris}(X/S)$. Then $\mathcal{F}$ comes equipped with a canonical integrable connection.

Proof. Say $(U, T, \delta )$ is an object of $\text{Cris}(X/S)$. Let $(U, T', \delta ')$ be the infinitesimal thickening of $T$ by $(\Omega _{X/S})_ T = \Omega _{T/S, \delta }$ constructed in Remark 60.13.1. It comes with projections $p_0, p_1 : T' \to T$ and a diagonal $i : T \to T'$. By assumption we get isomorphisms

$p_0^*\mathcal{F}_ T \xrightarrow {c_0} \mathcal{F}_{T'} \xleftarrow {c_1} p_1^*\mathcal{F}_ T$

of $\mathcal{O}_{T'}$-modules. Pulling $c = c_1^{-1} \circ c_0$ back to $T$ by $i$ we obtain the identity map of $\mathcal{F}_ T$. Hence if $s \in \Gamma (T, \mathcal{F}_ T)$ then $\nabla (s) = p_1^*s - c(p_0^*s)$ is a section of $p_1^*\mathcal{F}_ T$ which vanishes on pulling back by $i$. Hence $\nabla (s)$ is a section of

$\mathcal{F}_ T \otimes _{\mathcal{O}_ T} \Omega _{T/S, \delta }$

because this is the kernel of $p_1^*\mathcal{F}_ T \to \mathcal{F}_ T$ as $\mathcal{O}_{T'} = \mathcal{O}_ T \oplus \Omega _{T/S, \delta }$ by construction. It is easily verified that $\nabla (fs) = f\nabla (s) + s \otimes \text{d}(f)$ using the description of $\text{d}$ in Remark 60.13.1.

The collection of maps

$\nabla : \Gamma (T, \mathcal{F}_ T) \to \Gamma (T, \mathcal{F}_ T \otimes _{\mathcal{O}_ T} \Omega _{T/S, \delta })$

so obtained is functorial in $T$ because the construction of $T'$ is functorial in $T$. Hence we obtain a connection.

To show that the connection is integrable we consider the object $(U, T'', \delta '')$ constructed in Remark 60.13.2. Because $\mathcal{F}$ is a sheaf we see that

$\xymatrix{ q_0^*\mathcal{F}_ T \ar[rr]_{q_{01}^*c} \ar[rd]_{q_{02}^*c} & & q_1^*\mathcal{F}_ T \ar[ld]^{q_{12}^*c} \\ & q_2^*\mathcal{F}_ T }$

is a commutative diagram of $\mathcal{O}_{T''}$-modules. For $s \in \Gamma (T, \mathcal{F}_ T)$ we have $c(p_0^*s) = p_1^*s - \nabla (s)$. Write $\nabla (s) = \sum p_1^*s_ i \cdot \omega _ i$ where $s_ i$ is a local section of $\mathcal{F}_ T$ and $\omega _ i$ is a local section of $\Omega _{T/S, \delta }$. We think of $\omega _ i$ as a local section of the structure sheaf of $\mathcal{O}_{T'}$ and hence we write product instead of tensor product. On the one hand

\begin{align*} q_{12}^*c \circ q_{01}^*c(q_0^*s) & = q_{12}^*c(q_1^*s - \sum q_1^*s_ i \cdot q_{01}^*\omega _ i) \\ & = q_2^*s - \sum q_2^*s_ i \cdot q_{12}^*\omega _ i - \sum q_2^*s_ i \cdot q_{01}^*\omega _ i + \sum q_{12}^*\nabla (s_ i) \cdot q_{01}^*\omega _ i \end{align*}

and on the other hand

$q_{02}^*c(q_0^*s) = q_2^*s - \sum q_2^*s_ i \cdot q_{02}^*\omega _ i.$

From the formulae of Remark 60.13.2 we see that $q_{01}^*\omega _ i + q_{12}^*\omega _ i - q_{02}^*\omega _ i = \text{d}\omega _ i$. Hence the difference of the two expressions above is

$\sum q_2^*s_ i \cdot \text{d}\omega _ i - \sum q_{12}^*\nabla (s_ i) \cdot q_{01}^*\omega _ i$

Note that $q_{12}^*\omega \cdot q_{01}^*\omega ' = \omega ' \wedge \omega = - \omega \wedge \omega '$ by the definition of the multiplication on $\mathcal{O}_{T''}$. Thus the expression above is $\nabla ^2(s)$ viewed as a section of the subsheaf $\mathcal{F}_ T \otimes \Omega ^2_{T/S, \delta }$ of $q_2^*\mathcal{F}$. Hence we get the integrability condition. $\square$

Comment #1918 by Matthieu Romagny on

In the beginning of the proof, the diagonal is $i:T\to T'$ instead of $i:T\to T(1)$.

Comment #1919 by Matthieu Romagny on

In the proof, after the triangle diagram, write 'a commutative diagram' instead of 'a commutative map'.

Comment #5036 by MAO Zhouhang on

Seemingly the routine check that $\nabla$ is a connection is lacking. It seems better to at least mention the necessity of the verification.

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