Proposition 49.11.6. 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.

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

$\tilde\xi _ i : \Omega _{Y/S}^\bullet [-n_ i] \longrightarrow Rf_*\Omega ^\bullet _{X/S}$

in $D(Y, q^{-1}\mathcal{O}_ S)$. By the discussion in Cohomology, Section 20.31 on cohomology $\tilde\xi _ i$ 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 $D(Y, q^{-1}\mathcal{O}_ S)$. If $Y' \subset Y$ is open with inverse image $X' \subset X$, then $\tilde\xi |_{X'}$ induces the map

$\bigoplus \nolimits _{i = 1}^ N H^*_{dR}(Y'/S) \longrightarrow H^*_{dR}(X'/S), \quad (a_1, \ldots , a_ N) \longmapsto \sum f^*a_ i \cup \xi _ i|_{X'}$

on cohomology over $Y'$. Thus it suffices to find a basis for the topology on $Y$ such that the proposition holds for the members of the basis. This reduces us to the case discussed in the next paragraph.

Assume $Y$ and $S$ are affine. Say $Y = \mathop{\mathrm{Spec}}(A)$ and $S = \mathop{\mathrm{Spec}}(R)$. In this case $\Omega ^\bullet _{A/R}$ computes $R\Gamma (Y, \Omega ^\bullet _{Y/S})$ by Lemma 49.3.1. Choose a finite affine open covering $\mathcal{U} : X = \bigcup _{i \in I} U_ i$. Consider the complex

$K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/S}^\bullet ))$

as in Cohomology, Section 20.25. Let us collect some facts about this complex most of which can be found in the reference just given:

1. $K^\bullet$ is a complex of $R$-modules whose terms are $A$-modules,

2. $K^\bullet$ represents $R\Gamma (X, \Omega ^\bullet _{X/S})$ in $D(R)$ (Cohomology of Schemes, Lemma 29.2.2 and Cohomology, Lemma 20.25.2),

3. there is a natural map $\Omega ^\bullet _{A/R} \to K^\bullet$ of complexes of $R$-modules which is $A$-linear on terms and induces the pullback map $H^*_{dR}(Y/S) \to H^*_{dR}(X/S)$ on cohomology,

4. $K^\bullet$ has a multiplication denoted $\wedge$ which turns it into a differential graded $R$-algebra,

5. the multiplication on $K^\bullet$ induces the cup product on $H^*_{dR}(X/S)$ (Cohomology, Section 20.31),

6. the filtration $F$ on $\Omega ^*_{X/S}$ induces a filtration

$K^\bullet = F^0K^\bullet \supset F^1K^\bullet \supset F^2K^\bullet \supset \ldots$

by subcomplexes on $K^\bullet$ such that

1. $F^ kK^ n \subset K^ n$ is an $A$-submmodule,

2. $F^ kK^\bullet \wedge F^ lK^\bullet \subset F^{k + l}K^\bullet$,

3. $\text{gr}^ kK^\bullet$ is a complex of $A$-modules,

4. $\text{gr}^0K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/Y}^\bullet ))$ and represents $R\Gamma (X, \Omega ^\bullet _{X/Y})$ in $D(A)$,

5. multiplication induces an isomorphism $\Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0K^\bullet \to \text{gr}^ kK^\bullet$

We omit the detailed proofs of these statements; please see discussion leading up to the construction of the spectral sequence in Lemma 49.11.2.

For every $i = 1, \ldots , N$ we choose a cocycle $x_ i \in K^{n_ i}$ representing $\xi _ i$. Next, we look at the map of complexes

$\tilde x : M^\bullet = \bigoplus \nolimits _{i = 1, \ldots , N} \Omega ^\bullet _{A/R}[-n_ i] \longrightarrow K^\bullet$

which sends $\omega$ in the $i$th summand to $\omega \wedge x_ i$. All that remains is to show that this map is a quasi-isomorphism. We endow $M^\bullet$ with the structure of a filtered complex by the rule

$F^ kM^\bullet = \bigoplus \nolimits _{i = 1, \ldots , N} (\sigma _{\geq k}\Omega ^\bullet _{A/R})[-n_ i]$

With this choice the map $\tilde x$ is a morphism of filtered complexes. Observe that $\text{gr}^0M^\bullet = \bigoplus A[-n_ i]$ and multiplication induces an isomorphism $\Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0M^\bullet \to \text{gr}^ kM^\bullet$. By construction and Lemma 49.11.4 we see that

$\text{gr}^0\tilde x : \text{gr}^0M^\bullet \longrightarrow \text{gr}^0K^\bullet$

is an isomorphism in $D(A)$. It follows that for all $k \geq 0$ we obtain isomorphisms

$\text{gr}^ k \tilde x : \text{gr}^ kM^\bullet = \Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0M^\bullet \longrightarrow \Omega ^ k_{A/R}[-k] \otimes _ A \text{gr}^0K^\bullet = \text{gr}^ kK^\bullet$

in $D(A)$. Namely, the complex $\text{gr}^0K^\bullet = \text{Tot}(\check{\mathcal{C}}^\bullet (\mathcal{U}, \Omega _{X/Y}^\bullet ))$ is K-flat as a complex of $A$-modules by Lemma 49.11.5. Hence the tensor product on the right hand side is the derived tensor product as is true by inspection on the left hand side. Finally, taking the derived tensor product $\Omega ^ k_{A/R}[-k] \otimes _ A^\mathbf {L} -$ is a functor on $D(A)$ and therefore sends isomorphisms to isomorphisms. Arguing by induction on $k$ we deduce that

$\tilde x : M^\bullet /F^ kM^\bullet \to K^\bullet /F^ kK^\bullet$

is an isomorphism in $D(R)$ since we have the short exact sequences

$0 \to F^ kM^\bullet /F^{k + 1}M^\bullet \to M^\bullet /F^{k + 1}M^\bullet \to \text{gr}^ kM^\bullet \to 0$

and similarly for $K^\bullet$. This proves that $\tilde x$ is a quasi-isomorphism as the filtrations are finite in any given degree. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

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 input field. As a reminder, this is tag 0FMR. Beware of the difference between the letter 'O' and the digit '0'.