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Tag 07J6

Chapter 51: Crystalline Cohomology > Section 51.15: Connections

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}

    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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