# stacks/stacks-project

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
way...
 @@ -1101,6 +1101,8 @@ \section{de Rham cohomology of projective space} \section{The spectral sequence for a smooth morphism} \label{section-relative-spectral-sequence} @@ -1156,6 +1158,42 @@ \section{The spectral sequence for a smooth morphism} $$converging to R^{p + q}f_*\Omega^\bullet_{X/S}. \begin{remark} \label{remark-gauss-manin} In the situation above consider the cohomology sheaves$$ \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) \longrightarrow \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 $$with differentials given by d_1^{\bullet, q} is the de Rham complex of this connection. This connection is known as the {\it Gauss-Manin connection}. \end{remark} \begin{lemma} \label{lemma-cohomology-de-rham-base-change} Let f : X \to Y be a quasi-compact and quasi-separated morphism @@ -1257,6 +1295,164 @@ \section{The spectral sequence for a smooth morphism} the proof is finished. Some details omitted. \end{proof} \begin{proposition} \label{proposition-global-generation-on-fibres} Let f : X \to Y be a smooth proper morphism of schemes over a base S. Let N and n_1, \ldots, n_N \geq 0 be integers and let \xi_i \in H^{n_i}_{dR}(X/S), 1 \leq i \leq N. Assume for all points y \in Y the images of \xi_1, \ldots, \xi_N in H^*_{dR}(X_y/y) form a basis over \kappa(y). Then the map$$ \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. \end{proposition} \begin{proof} 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 \begin{align*} f^{-1}\Omega^\bullet_{Y/S}[-n_i] & = 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} \Omega^\bullet_{X/S} \end{align*} in D(X, p^{-1}\mathcal{O}_S). The adjoint of this is a map \tilde \xi_i : \Omega_{Y/S}^\bullet[-n_i] \longrightarrow Rf_*\Omega^\bullet_{X/S} $$which on cohomology (using the Leray spectral sequence) gives the map a \mapsto f^*a \cup \xi_i. Thus it suffices to show that the map$$ \tilde \xi = \bigoplus \tilde \xi_i : \bigoplus \Omega^\bullet_{Y/S}[-n_i] \longrightarrow Rf_*\Omega^\bullet_{X/S} $$is an isomorphism in D(Y, q^{-1}\mathcal{O}_S). \medskip\noindent Denote \overline{\xi}_i the image of \xi_i in H^{n_i}(Y, Rf_*\Omega^\bullet_{X/Y}). By Lemma \ref{lemma-proper-smooth-de-Rham} Rf_*\Omega^\bullet_{X/Y} is a perfect object of D(\mathcal{O}_Y) whose formation commutes with arbitrary base change. Thus the map$$ \bigoplus \overline{\xi}_i : \bigoplus \mathcal{O}_Y[-n_i] \longrightarrow Rf_*\Omega^\bullet_{X/Y} $$is an isomorphism in D(\mathcal{O}_Y) by our assumption on fibres. \medskip\noindent Recall that the spectral sequence discussed above comes from the filtration of \Omega^\bullet_{X/S} by the subcomplexes F^p\Omega^\bullet_{X/S} which are the image under \wedge of \text{Tot}( f^{-1}\sigma_{\geq p}\Omega^\bullet_{Y/S} \otimes_{f^{-1}\mathcal{O}_Y} \Omega^\bullet_{X/S}). Hence if we do the construction of \tilde \xi_i starting with the complex \sigma_{\geq p}\Omega^\bullet_{Y/S}, then we end up getting a morphism \tilde{\xi}_i^p : \sigma_{\geq p}\Omega^\bullet_{Y/S}[-n_i] \to Rf_*F^p\Omega^\bullet_{X/S}. These morphisms fit into the following commutative diagrams$$ \xymatrix{ \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} & Rf_*F^p\Omega^\bullet_{X/S} } $$and for p = 0 we obtain \tilde \xi_i. These diagrams imply that \tilde \xi_i^p induces maps$$ \overline{\xi}^p_i : \Omega^p_{Y/S}[-n_i] \longrightarrow 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} $$is an isomorphism. \medskip\noindent To finish the proof we would like to argue as follows. Consider \bigoplus \Omega_{Y/S}[-n_i] as a filtered complex on Y with filtration given by$$ \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 $$Our map \tilde \xi is a map of this filtered complex to the filtered complex Rf_*\Omega_{X/S} which induces an isomorphism on graded pieces. Hence it is an isomorphism in D(Y, q^{-1}\mathcal{O}_S). However, this isn't quite what we showed above (it would need a more precise discussion of maps between filtered complexes to show this). Instead we argue using cones as follows. We know that \bigoplus \overline{\xi}^p_i is an isomorphism for all p. Using Derived Categories, Lemma \ref{derived-lemma-third-isomorphism-triangle} we see that the induced map$$ \bigoplus \text{Cone}( \sigma_{\geq 2}\Omega^\bullet_{Y/S} \to \Omega^\bullet_{Y/S})[-n_i] \to \text{Cone}(Rf_*F^2\Omega^\bullet_{X/S} \to Rf_*\Omega^\bullet_{X/S}) $$is an isomorphism. Repeating we know that$$ \bigoplus \text{Cone}( \sigma_{\geq 3}\Omega^\bullet_{Y/S} \to \Omega^\bullet_{Y/S})[-n_i] \to \text{Cone}(Rf_*F^3\Omega^\bullet_{X/S} \to Rf_*\Omega^\bullet_{X/S})  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.) \end{proof}