## 50.13 Leray-Hirsch type theorems

In this section we prove that for a smooth proper morphism one can sometimes express the de Rham cohomology upstairs in terms of the de Rham cohomology downstairs.

Lemma 50.13.1. Let $f : X \to Y$ be a smooth proper morphism of schemes. Let $N$ and $n_1, \ldots , n_ N \geq 0$ be integers and let $\xi _ i \in H^{n_ i}_{dR}(X/Y)$, $1 \leq i \leq N$. 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

\[ \bigoplus \nolimits _{i = 1}^ N \mathcal{O}_ Y[-n_ i] \longrightarrow Rf_*\Omega ^\bullet _{X/Y} \]

associated to $\xi _1, \ldots , \xi _ N$ is an isomorphism.

**Proof.**
By Lemma 50.3.5 $Rf_*\Omega ^\bullet _{X/Y}$ is a perfect object of $D(\mathcal{O}_ Y)$ whose formation commutes with arbitrary base change. Thus the map of the lemma is a map $a : K \to L$ between perfect objects of $D(\mathcal{O}_ Y)$ whose derived restriction to any point is an isomorphism by our assumption on fibres. Then the cone $C$ on $a$ is a perfect object of $D(\mathcal{O}_ Y)$ (Cohomology, Lemma 20.49.7) whose derived restriction to any point is zero. It follows that $C$ is zero by More on Algebra, Lemma 15.75.7 and $a$ is an isomorphism. (This also uses Derived Categories of Schemes, Lemmas 36.3.5 and 36.10.7 to translate into algebra.)
$\square$

We first prove the main result of this section in the following special case.

Lemma 50.13.2. Let $f : X \to Y$ be a smooth proper morphism of schemes over a base $S$. Assume

$Y$ and $S$ are affine, and

there exist integers $N$ and $n_1, \ldots , n_ N \geq 0$ and $\xi _ i \in H^{n_ i}_{dR}(X/S)$, $1 \leq i \leq N$ such that 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

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

is an isomorphism.

**Proof.**
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 50.3.1. 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:

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

$K^\bullet $ represents $R\Gamma (X, \Omega ^\bullet _{X/S})$ in $D(R)$ (Cohomology of Schemes, Lemma 30.2.2 and Cohomology, Lemma 20.25.2),

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,

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

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

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

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

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

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

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

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 50.12.1.

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 $x_ i \wedge \omega $. 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 50.13.1 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 Derived Categories of Schemes, Lemma 36.23.3. 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.
$\square$

Proposition 50.13.3. 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 $\kappa (y)$. The map

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

(see proof) is an isomorphism in $D(Y, (Y \to S)^{-1}\mathcal{O}_ S)$ and correspondingly the map

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

is an isomorphism.

**Proof.**
Denote $p : X \to S$ and $q : Y \to S$ be the structure morphisms. Let $\xi '_ i : \Omega ^\bullet _{X/S} \to \Omega ^\bullet _{X/S}[n_ i]$ be the map of Remark 50.4.3 corresponding to $\xi _ i$. Denote

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

the composition of $\xi '_ i$ with the canonical map $\Omega ^\bullet _{Y/S} \to Rf_*\Omega ^\bullet _{X/S}$. Using

\[ R\Gamma (Y, Rf_*\Omega ^\bullet _{X/S}) = R\Gamma (X, \Omega ^\bullet _{X/S}) \]

on cohomology $\tilde\xi _ i$ is the map $\eta \mapsto \xi _ i \cup f^*\eta $ from $H^ m_{dR}(Y/S)$ to $H^{m + n}_{dR}(X/S)$. Further, since the formation of $\xi '_ i$ commutes with restrictions to opens, so does the formation of $\tilde\xi _ i$ commute with restriction to opens.

Thus we can consider the map

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

To prove the lemma it suffices to show that this is an isomorphism in $D(Y, q^{-1}\mathcal{O}_ S)$. If we could show $\tilde\xi $ comes from a map of filtered complexes (with suitable filtrations), then we could appeal to the spectral sequence of Lemma 50.12.1 to finish the proof. This takes more work than is necessary and instead our approach will be to reduce to the affine case (whose proof does in some sense use the spectral sequence).

Indeed, if $Y' \subset Y$ is is any open with inverse image $X' \subset X$, then $\tilde\xi |_{X'}$ induces the map

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

on cohomology over $Y'$, see discussion above. Thus it suffices to find a basis for the topology on $Y$ such that the proposition holds for the members of the basis (in particular we can forget about the map $\tilde\xi $ when we do this). This reduces us to the case where $Y$ and $S$ are affine which is handled by Lemma 50.13.2 and the proof is complete.
$\square$

## Comments (0)