The Stacks project

Lemma 59.15.1. In Situation 59.7.5. 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 59.13.1. 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 59.13.1.

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 59.13.2. 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 59.13.2 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. $\square$


Comments (5)

Comment #1918 by Matthieu Romagny on

In the beginning of the proof, the diagonal is instead of .

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 is a connection is lacking. It seems better to at least mention the necessity of the verification.

There are also:

  • 2 comment(s) on Section 59.15: Connections

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07J6. Beware of the difference between the letter 'O' and the digit '0'.