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.

## 60.15 Connections

In Situation 60.7.5. Given an $\mathcal{O}_{X/S}$-module $\mathcal{F}$ on $\text{Cris}(X/S)$ a *connection* is a map of abelian sheaves

such that $\nabla (f s) = f\nabla (s) + s \otimes \text{d}f$ for local sections $s, f$ of $\mathcal{F}$ and $\mathcal{O}_{X/S}$. Given a connection there are canonical maps $ \nabla : \mathcal{F} \otimes _{\mathcal{O}_{X/S}} \Omega ^ i_{X/S} \longrightarrow \mathcal{F} \otimes _{\mathcal{O}_{X/S}} \Omega ^{i + 1}_{X/S} $ defined by the rule $\nabla (s \otimes \omega ) = \nabla (s) \wedge \omega + s \otimes \text{d}\omega $ as in Remark 60.6.8. We say the connection is *integrable* if $\nabla \circ \nabla = 0$. If $\nabla $ is integrable we obtain the *de Rham complex*

on $\text{Cris}(X/S)$. It turns out that any crystal in $\mathcal{O}_{X/S}$-modules 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

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

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

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

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

and on the other hand

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

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. $\square$

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## Comments (2)

Comment #4172 by Zeyu Liu on

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