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
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
Comments (5)
Comment #1918 by Matthieu Romagny on
Comment #1919 by Matthieu Romagny on
Comment #1988 by Johan on
Comment #5036 by MAO Zhouhang on
Comment #5265 by Johan on
There are also: