# The Stacks Project

## Tag 07J6

Lemma 54.15.1. In Situation 54.7.5. 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 54.13.1. 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 54.13.2. 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 54.13.2 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}

## Comments (3)

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.

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