# The Stacks Project

## Tag 07J6

Lemma 51.15.1. In Situation 51.7.5. 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 51.13.1. 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 $\Delta$. 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 $\Omega_{T/S, \delta}$ is the kernel of $\mathcal{O}_{T'} \to \mathcal{O}_T$ by construction.

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 51.13.2. 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 51.13.2 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. $\square$

The code snippet corresponding to this tag is a part of the file crystalline.tex and is located in lines 2684–2690 (see updates for more information).

\begin{lemma}
\label{lemma-automatic-connection}
In Situation \ref{situation-global}.
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.
\end{lemma}

\begin{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 \ref{remark-first-order-thickening}.
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 $\Delta$. 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 $\Omega_{T/S, \delta}$ is the kernel of
$\mathcal{O}_{T'} \to \mathcal{O}_T$ by construction.

\medskip\noindent
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.

\medskip\noindent
To show that the connection is integrable we consider the
object $(U, T'', \delta'')$ constructed in
Remark \ref{remark-second-order-thickening}.
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 \ref{remark-second-order-thickening} 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.
\end{proof}

Comment #1918 by Matthieu Romagny on April 16, 2016 a 12:58 pm UTC

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 April 16, 2016 a 12:59 pm UTC

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

Comment #1988 by Johan (site) on May 3, 2016 a 1:51 pm UTC

OK, thanks! I fixed both issues here.

## Add a comment on tag 07J6

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 lower-right corner).

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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?