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Global classes generating over fibres

In the case we are going to apply this, we could work locally on Y and
then we could use the Kunneth formula... but it is a better result this
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aisejohan committed Sep 27, 2019
1 parent cff9d55 commit b78a66a7c8b9f4aae44da9c095b60c0a74586276
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  1. +196 −0 derham.tex
@@ -1101,6 +1101,8 @@ \section{de Rham cohomology of projective space}

\section{The spectral sequence for a smooth morphism}

@@ -1156,6 +1158,42 @@ \section{The spectral sequence for a smooth morphism}
converging to $R^{p + q}f_*\Omega^\bullet_{X/S}$.

In the situation above consider the cohomology sheaves
\mathcal{H}^q_{dR}(X/Y) = H^q(Rf_*\Omega^\bullet_{X/Y}))
If $f$ is proper in addition to being smooth and $S$ is a scheme over
$\mathbf{Q}$ then $\mathcal{H}^q_{dR}(X/Y)$ is finite locally free (insert
future reference here). If we only assume $\mathcal{H}^q_{dR}(X/Y)$
are flat $\mathcal{O}_Y$-modules, then we obtain (tiny argument omitted)
E_1^{p, q} =
\Omega^p_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y)
and the differentials in the spectral sequence are maps
d_1^{p, q} :
\Omega^p_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y)
\Omega^{p + 1}_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y)
In particular, for $p = 0$ we obtain a map
$d_1^{0, q} : \mathcal{H}^q_{dR}(X/Y) \to
\Omega^1_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y)$
which turns out to be an integrable connection (insert future reference here)
and the complex
\mathcal{H}^q_{dR}(X/Y) \to
\Omega^1_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y) \to
\Omega^2_{Y/S} \otimes_{\mathcal{O}_Y} \mathcal{H}^q_{dR}(X/Y) \to \ldots
with differentials given by $d_1^{\bullet, q}$
is the de Rham complex of this connection.
This connection is known as the {\it Gauss-Manin connection}.

Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism
@@ -1257,6 +1295,164 @@ \section{The spectral sequence for a smooth morphism}
the proof is finished. Some details omitted.

Let $f : X \to Y$ be a smooth proper morphism of schemes over a base $S$.
Let $N$ and $n_1, \ldots, n_N \geq 0$ be integers and let
$\xi_i \in H^{n_i}_{dR}(X/S)$, $1 \leq i \leq N$.
Assume for all points $y \in Y$ the images of $\xi_1, \ldots, \xi_N$
in $H^*_{dR}(X_y/y)$ form a basis over $\kappa(y)$. Then the map
\bigoplus\nolimits_{i = 1}^N H^*_{dR}(Y/S) \longrightarrow
H^*_{dR}(X/S), \quad
(a_1, \ldots, a_N) \longmapsto \sum f^*a_i \cup \xi_i
is an isomorphism.

Denote $p : X \to S$ and $q : Y \to S$ be the structure morphisms.
We can think of
$\xi_i$ as a map $p^{-1}\mathcal{O}_S[-n_i] \to \Omega^\bullet_{X/S}$.
Thus we can consider the map
& =
f^{-1}\Omega^\bullet_{Y/S} \otimes_{p^{-1}\mathcal{O}_S}^\mathbf{L}
p^{-1}\mathcal{O}_S[-n_i] \\
& \xrightarrow{\text{id} \otimes \xi_i}
f^{-1}\Omega^\bullet_{Y/S} \otimes_{p^{-1}\mathcal{O}_S}^\mathbf{L}
\Omega^\bullet_{X/S} \\
& \to
\text{Tot}(f^{-1}\Omega^\bullet_{Y/S} \otimes_{p^{-1}\mathcal{O}_S}
\Omega^\bullet_{X/S}) \\
& \xrightarrow{\wedge}
in $D(X, p^{-1}\mathcal{O}_S)$. The adjoint of this is a map
\tilde \xi_i :
which on cohomology (using the Leray spectral sequence)
gives the map $a \mapsto f^*a \cup \xi_i$. Thus it suffices
to show that the map
\tilde \xi = \bigoplus \tilde \xi_i :
\bigoplus \Omega^\bullet_{Y/S}[-n_i]
is an isomorphism in $D(Y, q^{-1}\mathcal{O}_S)$.

Denote $\overline{\xi}_i$ the image of $\xi_i$ in
$H^{n_i}(Y, Rf_*\Omega^\bullet_{X/Y})$.
By Lemma \ref{lemma-proper-smooth-de-Rham}
$Rf_*\Omega^\bullet_{X/Y}$ is a perfect object of $D(\mathcal{O}_Y)$
whose formation commutes with arbitrary base change.
Thus the map
\bigoplus \overline{\xi}_i :
\bigoplus \mathcal{O}_Y[-n_i]
is an isomorphism in $D(\mathcal{O}_Y)$ by our assumption on fibres.

Recall that the spectral sequence discussed above comes from the
filtration of $\Omega^\bullet_{X/S}$ by the subcomplexes
$F^p\Omega^\bullet_{X/S}$ which are the image under $\wedge$ of
f^{-1}\sigma_{\geq p}\Omega^\bullet_{Y/S} \otimes_{f^{-1}\mathcal{O}_Y}
Hence if we do the construction of $\tilde \xi_i$ starting with the complex
$\sigma_{\geq p}\Omega^\bullet_{Y/S}$, then we end up getting a morphism
$\tilde{\xi}_i^p : \sigma_{\geq p}\Omega^\bullet_{Y/S}[-n_i] \to
Rf_*F^p\Omega^\bullet_{X/S}$. These morphisms
fit into the following commutative diagrams
\sigma_{\geq p + 1}\Omega_{Y/S}^\bullet[-n_i]
\ar[r]_{\tilde \xi^{p + 1}_i} \ar[d] &
Rf_*F^{p + 1}\Omega^\bullet_{X/S} \ar[d] \\
\sigma_{\geq p}\Omega_{Y/S}^\bullet[-n_i] \ar[r]^{\tilde \xi^p_i} &
and for $p = 0$ we obtain $\tilde \xi_i$. These diagrams imply that
$\tilde \xi_i^p$ induces maps
\overline{\xi}^p_i :
Rf_*\text{gr}^p\Omega^\bullet_{X/S} =
\Omega^p_{Y/S}[-p] \otimes_{\mathcal{O}_Y}^\mathbf{L} Rf_*\Omega_{Y/X}
where the equal sign comes from the discussion above (more precisely
from Lemma \ref{lemma-cohomology-de-rham-base-change}).
Since we have seen above that $\bigoplus \overline{\xi}_i$
is an isomorphism and since $\overline{\xi}^p_i$ is compatible
with $\overline{\xi}_i$ (details omitted), we conclude that
\bigoplus \overline{\xi}^p_i :
\bigoplus \Omega^p_{Y/S}[-n_i] \to
\Omega^p_{Y/S}[-p] \otimes_{\mathcal{O}_Y}^\mathbf{L} Rf_*\Omega_{Y/X}
is an isomorphism.

To finish the proof we would like to argue as follows. Consider
$\bigoplus \Omega_{Y/S}[-n_i]$
as a filtered complex on $Y$ with filtration given by
\bigoplus \Omega_{Y/S}[-n_i] \supset
\bigoplus \sigma_{\geq 1}\Omega_{Y/S}[-n_i] \supset
\bigoplus \sigma_{\geq 2}\Omega_{Y/S}[-n_i] \supset \ldots
Our map $\tilde \xi$ is a map of this filtered complex
to the filtered complex $Rf_*\Omega_{X/S}$ which induces
an isomorphism on graded pieces. Hence it is an isomorphism
in $D(Y, q^{-1}\mathcal{O}_S)$. However, this isn't quite what
we showed above (it would need a more precise discussion of
maps between filtered complexes to show this).
Instead we argue using cones as follows. We know that
$\bigoplus \overline{\xi}^p_i$ is an isomorphism for all $p$.
Using Derived Categories, Lemma \ref{derived-lemma-third-isomorphism-triangle}
we see that the induced map
\sigma_{\geq 2}\Omega^\bullet_{Y/S}
\text{Cone}(Rf_*F^2\Omega^\bullet_{X/S} \to
is an isomorphism. Repeating we know that
\sigma_{\geq 3}\Omega^\bullet_{Y/S}
\text{Cone}(Rf_*F^3\Omega^\bullet_{X/S} \to
is an isomorphism. And so on. Now if we want to check that $\tilde \xi$
is an isomorphism in cohomological degree $n$, then
it suffices to look at the $n$th cone in this process
and we get the conclusion. (If $Y$ is of
finite type over $S$, then $\sigma_{\geq n}\Omega^\bullet_{Y/S}$
is zero for some $n$ and hence we get the isomorphism directly
in this case.)

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